Practice: A block of unknown mass is initially at rest on a frictionless horizontal surface. When you push on it with a constant horizontal force of 5 N, the block starts to move and covers 24 m in the first 6 seconds. Find the mass of the block.

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Sections | |||
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Intro to Forces + Newton's Laws | 16 mins | 0 completed | Learn |

Force Problems with Motion | 28 mins | 0 completed | Learn |

Forces with Multiple Objects | 26 mins | 0 completed | Learn |

Vertical Forces & Equilibrium | 34 mins | 0 completed | Learn |

More 1D Equilibrium | 16 mins | 0 completed | Learn |

Vertical Forces & Acceleration | 26 mins | 0 completed | Learn |

Landing & Jumping Problems | 19 mins | 0 completed | Learn |

Forces in 2D | 35 mins | 0 completed | Learn |

2D Equilibrium | 25 mins | 0 completed | Learn |

Concept #1: Force Problems with Motion

**Transcript**

Hey guys. So now that I've introduced forces as well as shown you the equation, F equals ma, it's time to look at how some force problems work. The first type of problems we're going to look into actually have to do with motion kinematics. So we're going to get those three to four equations we used to work with a lot and combine that with F equals ma, so let me show you how it works. So Force problems with motion is what I call this and it says your some force problems will make or some problems mix forces with motion. We're going to use F equals ma the equations of motion, here the equations of motion they four of them. So of you might only be allowed to use three of them remember. OK. So remember UAM equations that's what I call them. They have five variables. Vinitial, Vfinal Delta x now the T and A. And if you look at F these are the motion variables if you look F equals ma well has F M And the reason why I do it like this explain that a exists in all in all of these equations right. So not only motion equations and variables but also in force in the force equation F equals, So you can imagine what can happen in motion problems you're going to need two I'm sorry, three out of five variables to be able to solve a problem. But now instead what I might do or what a question might do is give you these two not give you a third but then give you these two. So by giving you F and m you're going to be able to find a. Which means you're now going to have three or five here. So you just have to jump around from one to the other. It's pretty straightforward let's do some examples here. So it says a 20 kilogram block accelerates from rest to 30 meters per second in six seconds. So we draw a 20 kilogram block. It accelerates so there must be a force pushing it because I know from F equals ma that if I have an acceleration I have to have a force and it's being pulled let's say in that way ahe goes from an initial velocity of zero to a velocity of 30 in just a time of just six seconds. And what we to know what is the average force exerted on the block. Now don't worry that it's an average can be thought of as a constant force. So it's just F right. And how do I find F. Well the sum of all forces on an object equals ma. The only force here is this force pushing forward plus F the mass of this thing is 20 and acceleration. Now I'm looking for F so I have to have acceleration but I'm obviously not going to get it out of this equation because there's two unknowns So what I'm going to do I'm going to go to motion to my three equations of motion to try to find acceleration. OK. So I already have one, two, three. I have enough variables to figure out acceleration. The third variable the fifth variable here is delta x and that variable is my ignored variable so I put a little sad face here. So I should use the equation. Remember that you that doesn't have my ignored variable delta x and that is the first equation. So to find acceleration I can go here. The final velocity is 30, the initial velocity zero, acceleration is what I'm looking for and time is six. If I move things around 30 over 6 or simply 5 meters per second square. Now that I got the acceleration I can just go back in here, F equals twenty a or twenty times five so the average force is one hundred and that's the final answer. OK. Let's do another one. A five kilogram block is initially at rest on the frictionless horizontal surface. So you have a five kilogram block. It's initially at rest so the initial velocity here is zero. And you push on it with a constant force of 10 and you do this for four seconds. So you can kind of draw a little interval here and say that this takes time of four seconds and I want to know the blocks displacement and the blocks speed after four seconds. OK let's do one at a time. The first question is what is the displacement or what is delta x.

This is a motion variable delta x doesn't belong in F equals ma it's not there so I'm gonna have to use motion equations to figure this out and to use motion and have to have three or five variables but I only have two, the first variable that I'm missing here is the acceleration. So we have two variables. So I need a third one and that third one cannot be delta x because that's what I'm looking for. The third one is going to be a or v but if you notice I'm asking for v later and I wouldn't have a place to find v anyway. So this variable is going to be my a and I'm going to get a from F equals ma. So let's do that. The sum of all forces equals ma. The only force here is a positive 10 because it's to the right. It doesn't say that it's the right thing it's hard so I'm just gonna stick with it, a mass equals five and acceleration is what I'm looking for it's 10 over 5 acceleration is 2 meters per second square. OK. So this acceleration here I know is 2. Now I know three out of five variables and I'm ready to find out to. This v I'm going to want it later. But for now it's my ignored variables this final velocity here which tells me that I should use the third equation delta x equals Vinitial T plus half of at square. This goes away and I'm left with just this. So the final answer is 16 meters for the first part. All right for Part B I want to find the final velocity part B final velocity. So part B is this guy right here. And now this is just a motion problem except that I know four variables I know that this delta x is 16. So I can just pick whichever equation is easiest the first equations easiest V equals Vinitial plus at, V equals zero plus 2 times four this is pretty straightforward. Eight meters per second. Call. All right so I want you guys to try this one out. And it's very similar to the ones that I have done except that I'm looking for the mass of the block. So give us a shot.

Practice: A block of unknown mass is initially at rest on a frictionless horizontal surface. When you push on it with a constant horizontal force of 5 N, the block starts to move and covers 24 m in the first 6 seconds. Find the mass of the block.

Example #1: More Forces with Motion

**Transcript**

Hey guys so let's do a few more force problems involving motion Let's get started. So in this example I have two skaters and they're dressed in frictionless ice and they're going to push each other, A pushes B with us force of two hundred towards the end positive X. axis so positive X. axis also this way if A pushes B that way than a should be on the left so I'm going to make a girl and B going to be a dude. These are suppose to be skates that's why their are long feet. All right so she's pushing him so this is she's A so mass of A is forty and it's B mass of B is fifty and she's pushing him. Force of A on B. which I can draw is a push but I would rather draw it as a pull, okay and she when she pushes on him action reaction he pushes in her as well and that force has to have the same magnitude just opposite directions F of B on A it's a mutual push and every push is a mutual push. So if this is two hundred then I know that this is two hundred as well and in fact that's the first question the first question is what is the average force that B exerts on A and the answer is simply two hundred. To the left. I could have said negative two hundred. Since I said they're going to the right is positive so this forces positive this force is negative again and I have to put the signs there because I have the arrows show the direction that's what it is and I want to know the average speeds, I'm sorry the speeds of A and B after the point five seconds that they pushed each other for OK so I will know for a part b I wanna know the final velocity of A. So I can think of this as a motion problem the initial velocity of A is zero the acceleration of a. I don't know yet. What else is either Delta X. I don't know and deltaT is point five seconds OK this is a motion problem because initial velocity or any of the any of the two losses is going to come out of the motion equation not out of F equals ma, right so if you notice here I have one too and I'm trying to find the final velocity so either one of these two guys is going to have to be my third variable so I can solve this we've done this a bunch of times and you know that it's going to be your airy coming from F. it was a mason so I gave you the force in the masses OK So in this part is really important when you write F equals ma you write F equals ma for an object and you write it for the object that has the forces acting on it so force is always by an object on all your objects when you write the sum of our forces the sum of all forces on an object and then this is the force of the object that has the forces acting I'm sorry this is the mass of the object that has the forces acting on it OK and this is the acceleration of the object so the forces on the masses of and the acceleration of OK so some more force is on if I want to the acceleration of a I'm going to use the forces only in the mass of a. Plane So what are the forces on A well A is is the girl the girl being pushed to the left with two hundred so the force is the only force on hers. Two hundred to the left her mass because the sum of all forces on the girl the mass all of the girl forty and this gives us the acceleration of the girl OK OK And if you do this the Excel ration is two and by the forty there's a negative the negative five meters per second squared OK. So this is a five now I can put this five in here I know three out of five things this is my ignore variable and this tells me I can use the first equation. So the five The last is negative two point five meters per second but boom right so the other one is to find a final velocity of be going to go to the same process the initial velocity of be zero the acceleration of the I don't have Delta X. I don't know it's a mutual pushing the mutual push each other for point five B. announces exactly the same and I need to find A using F equals ma so the sum of all forces on B. is the mass of B. and the acceleration of the ok B as if you look at B. B.'s the deal over here he's been pushed to the right with two hundred. His mass is a fifty and his acceleration is what I'm looking for so his acceleration is going to be positive for meters per second squared I can put this four in here and now again I can use equation number one to figure this out. So this is zero plus positive for. Or point five So the last is true meters per second now notice that her velocity is negative his velocity is positive that should make sense she is pushing him to the right sir she ends up going to the left and then the same thing with him has been pushed to the right to go to the right.

OK I notice that the forces are the same. Because of action reaction but the accelerations are not OK if you call them A The forces are the same but the masses are different so the accelerations will be different OK let me do one more question and then I want you guys to do some of them. They fifteen kilogram crates movie you want to hear is all to surface with twelve meters per second so here's a fifteen kilogram creates. This by the way is not a free body diagram remember everybody diagram would be just a dart with forces coming out of it if you draw a little box it's schematics for the problem not everybody diagram but anyway it's moving with a velocity of twelve and then it takes four seconds to come to a stop so coming to a stop means your final velocity is zero. In if you take four seconds that means a delta T. equals four cool. It does this due to friction with the surface now we have introduce friction here that's going to come later but I can tell you already that friction and you kind of know this already friction opposes motion so if this box is moving this way with twelve meters per second friction is this way OK And in this case the only force them is object is friction the free body diagram would look like this. A lot of people feel like they should be riding a force going forward as well because the object is moving twelve or remember. It's moving with twelve because it only had twelve meters per second because at some point someone pushed or whatever doesn't matter you don't need a force to have that twelve to continue or twelve ok. so the only force here is friction OK therefore we should expect to look at the acceleration is to the left F equals ma if the only force left the object should, accelerates the left OK And I want to know what is the force of friction so friction is a force in the way I'm going to fight forces calculate a force and a force problem is using half it was in May some other forces on the block mass of a block acceleration of the block the only force is friction friction is going to have to see the going to the right is positive so friction is negative so when I do have it was a ma I plug in forces with their respective with their appropriate sine masses fifteen and acceleration zero have now for me to find friction I'm kind of stuck here I'm going to have to find a way and that's going to come from motion problem right from the motion equations so if I want to find a. Notice that I already have three variables so this is actually pretty easy I can just use the first equation the final the initial plus eighty the final the last twelve and so the final loss is actually zero the initial loss is twelve the accelerations I'm looking for in the Times for the twelve goes over the years a negative.

And looks like this so the acceleration is negative three it should make sense that I have a negative acceleration because if my only four still left my acceleration should be to the left should be negative OK now I can plug this in here in knows that this remains negative friction fifty mm negative three This means friction equals forty five Newtons and I want to remind you that even though I got a positive Acer. It doesn't mean the frictions to the right again you can plug in forces here with their correct sign indicating direction and then when F. it was a maze going to give you as the final answer is the magnitude I know it's negative that's why I made a native there OK in the fact that I made a negative actually in the cancelling that negative later on so just remember the answer you get here plug ins are negative the answer you get here is your magnitude OK So friction in this case is forty five Newton's. All right so now I want you guys to try these two practice problems. There are very similar to questions that I've done before and I thought you got a right to try.

Practice: A 1,000-kg car leaves a skid mark of 80 m while coming to a stop. If the maximum force the brakes are capable of is 8,000N, find the car’s initial velocity before braking.

Practice: A gun shoots a 10 g bullet out of its 8.0 cm-long barrel with a muzzle speed of 400 m/s. Find the force applied on the bullet by the gun. Find force (magnitude and direction) applied on the gun by the bullet.

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Concept #1: Force Problems with Motion

Practice

Example #1: More Forces with Motion

Practice

Practice

Under the influence of a constant force, an object, starting from rest, moves a distance d in time t. Under the influence of the same force, how far will the object move in a time 2t, again starting from rest?A) 2dB) 4dC) sqrt (2) dD) the same distance d

True or False: a) In the Newton's law of gravitation, F = Gm1m2/r2, the masses m1 and m2 refer to inertial masses. b) For a cart moving along a track, with the x-axis parallel to the track, only the x-coordinate of the cart can change. c) If the cart is moving on the positive x axis, the x-component of velocity must be positive.

Scallops eject water from their shells to provide a thrust force. (Figure 1) shows a smoothed graph of actual data for the initial motion of a 25g scallop speeding up to escape a predator.What is the magnitude of the net force needed to achieve this motion?Express your answer to two significant figures and include the appropriate units.

You've just kicked a rock on the sidewalk and it is now sliding along the concrete. (A) Which of the motion diagrams shown in the figure below is correct system?

A ball is shot from a compressed air gun at twice its terminal speed.a)What is the ball's initial acceleration, as a multiple of g, if it is shot straight up?b) What is the ball's initial acceleration, as a multiple of g, if it is shot straight down?

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion in 24 N, what force F (in newtons) is the person exerting on the mower?

A jet plane is speeding down the runway during takeoff. Air resistance is not negligible, friction is negligible.Identify the forces on the jet. Choose your answers from the following list:a) Gravityb) Spring Forcec) Tensiond) Normal Forcee) Frictionf) Dragg) Thrust

A 1500 kg rocket takes off vertically from a launch pad. Its engines provide a thrust of 25,000 N. After 20.0 s, what is the speed of the rocket?A) 328 m/sB) 78 m/sC) 112 m/sD) 137 m/sE) 16 m/s

A brave but inadequate rugby player is being pushed backward by an opposing player who is exerting a force of 800 N on him. The mass of the losing player plus equipment is 90.0 kg, and he is accelerating at 120 m/s2 backward. (a) What is the force of friction between the losing player’s feet and the grass? (b) What force does the winning player exert on the ground to move forward if his mass plus equipment is 110 kg? (c) Draw a sketch of the situation showing the system of interest used to solve each part. For this situation, draw a free-body diagram and write the net force equation.

A helicopter of mass M is lowering a truck of mass m onto the deck of a ship, (a) At first, the helicopter and the truck move downward together (the length of the cable doesn't change). If their downward speed is decreasing at a rate of 0.467 g, what is the tension in the cable? T = ______ mg (b) As the truck gets close to the deck, the helicopter stops moving downward. While it hovers, it lets out the cable so that the truck is still moving downward. If the truck's downward speed is decreasing at a rate of 0.467 g, while the helicopter is at rest, what is the tension in the cable? T = ______ mg

The acceleration of a baseball pitcher's hand as he delivers a pitch is extreme. For a professional player, this acceleration phase lasts only 50 ms, during which the ball's speed increases from 0 to about 90 mph, or 40 m/s.What is the force of the pitcher's hand on the 0.145 kg ball during this acceleration phase? Express your answer with the appropriate units.

A gymnast of mass 69.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s2 for the acceleration of gravity.Calculate in Newtons:The tension T in the rope if the gymnast hangs motionless on the rope.

A gymnast of mass 69.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s2 for the acceleration of gravity.Calculate in Newtons:The tension T in the rope if the gymnast climbs the rope at a constant rate.

A gymnast of mass 69.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s2 for the acceleration of gravity.Calculate in Newtons:The tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 0.700 m/s2.

A gymnast of mass 69.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s2 for the acceleration of gravity.Calculate the tension T in the rope (in newtons) if the gymnast slides down the rope with a downward acceleration of magnitude 0.700 m/s2.

To understand the concept of tension and the relationship between tension and force. This problem introduces the concept of tension. The example is a rope, oriented vertically, that is being pulled from both ends. (Figure 1) Let Fu and Fd (with u for up and d for down) represent the magnitude of the forces acting on the top and bottom of the rope, respectively. Assume that the rope is massless so that its weight is negligible compared with the tension. (This is not a ridiculous approximation-modern rope materials such as Kevlar can carry tensions thousands of times greater than the weight of tens of meters of such rope.) Consider the three sections of rope labeled a, b, and c in the figure. At point 1, a downward force of magnitude Fad acts on section a. At point 1, an upward force of magnitude Fbu acts on section b. At point 1, the tension in the rope is T1. At point 2, a downward force of magnitude Fbd acts on section b. At point 2, an upward force of magnitude Fcu acts on section c. At point 2, the tension in the rope is T2. Assume, too, that the rope is at equilibrium.Part AWhat is the magnitude Fad of the downward force on section a?Express your answer in terms of the tension T1.Part BWhat is the magnitude Fbu of the upward force on section b?Express your answer in terms of the tension T1.Part CThe magnitude of the upward force on c, Fcu, and the magnitude of the downward force on b, Fbd, are equal because of which of Newton's laws?Part DThe magnitude of the force Fbu is ____ Fbd.Part ENow consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces?Now consider the forces on the ends of the rope. What is the relationship between the magnitudes of these two forces?a) Fu > Fdb) Fu < Fdc) Fu = FdPart FThe ends of a massless rope are attached to two stationary objects (e.g., two trees or two cars) so that the rope makes a straight line. For this situation, which of the following statements are true?Check all that apply.a) The tension in the rope is everywhere the same.b) The magnitudes of the forces exerted on the two objects by the rope are the same.c) The forces exerted on the two objects by the rope must be in opposite directions.d) The forces exerted on the two objects by the rope must be in the direction of the rope

You've just kicked a rock on the sidewalk and it is now sliding along the concrete. Draw a free body diagramDraw the force vectors with their tails at the dot. The orientation of your vectors will be graded. The exact length of your vectors will not be graded but the relative length of one to the other will be graded.

In Figure 4.7, the net external force on the 24-kg mower is stated to be 51 N. The force of friction opposing the motion in 24 N. Suppose the mower is moving at 1.5 m/s when the force F is removed. How far will the mower go before stopping?

If you experience weightlessness, this means that:A. There is no net force acting on you.B. You are accelerating upward.C. Your apparent weight is zero.

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