(b) Doug uses a 25 N horizontal force to push a 5.0 kg crate up a 2.0-m-high, 20° frictionless slope. What is the speed of the crate at the top of the slope?

Question

(b) Doug uses a 25 N horizontal force to push a 5.0 kg crate up a 2.0-m-high, 20° frictionless slope. What is the speed of the crate at the top of the slope?

Accepted Answer

Hey, everyone. So this problem is dealing with work. Let's see what they're asking us. A storekeeper pushes a 15 kg trolley initially at rest of a three m long ramp inclined at an angle of degrees. If the storekeeper applies a force of 120 newtons horizontally, determine the speed of the trolley as it exits the ramp. So our multiple choice answers here are a 2. m per second. B 4.32 m per second. C 9.84 m per second or D 4.92 m per second. So the first thing we're going to do here is to draw a diagram of what is happening in this problem. So we had a rat, we're told that it's angle at 25 degrees. We have the trolley that weighs 15 kg. And so the forces acting on this trolley are going to be weight in the negative direction. And I'm not saying negative Y here. So we'll talk about that in a sec. So we in the direction that is perpendicular to the ground, we'll say OK, the normal force is in the positive direction that is perpendicular to the ramp surface. And then we have the force of the storekeeper which is apply horizontal two, the, the ground surface. So I call that the pushing force. OK. So we can determine if we want to use the kind of traditional X and Y axis where X would be the ground or we can determine we can use our own X and Y axis. And I'm going to, you know, X as being the direction of the ram. So X is going to be in the direction of the ramp and then the positive Y is going to be perpendicular two, the right. OK. So now let's go back to the problem. They're asking for the speed of the trolley and they are giving us a distance and a force. So we can, we can recall two different equations here. The first is that work by definition is force times the displacement. We can also recall from our work energy theorem that the net work is given by the change in kinetic energy. And in turn kinetic energy is given as one off and B square. So paired with Newton's second law with using our diagram of our forces, we can find this seed here in the kinetic energy equation. So the next step is to take the sum of the forces in the X direction. So we can recall from Newton's second law of some of the forces is equal to mass multiplied by acceleration. And in this case, we're looking for some of the forces in the X direction. So some of the forces in the X direction are going to be our p our pushing force multiplied by the cosine of 25. Because the X component of that pushing force is um that the pushing force multiplied by the cosine of 25. And then for the weight force, we're looking for the X component of the weight force. And that's gonna be weight multiplied by sine of 25. And how we have defined our X and Y coordinates, there is no X component of the normal force. So our, some of the, some of the forces in the X direction, we are told that in the problem P is 100 and 20 newtons. So we can flood that in. So we have 120 newtons multiplied by the cosine and 25. And then our weight we can recall is given by mass multiplied by gravity. So the mass was given in the problem as 15 kg. Gravity is 9.8 m per second squared. And then that multiplied by the sign of 25. We need units over here. So that's 100 and 20 units. So sorry for that. But we can plug this into our calculator and we get the net force in the X direction is equal to 46.6 students. And now our work is given by force multiplied by displacement, we can plug in that force that we just found. So 46.6 newtons at our displacement is this three m along the ramp. So force multiplied by displacement, we plug that into our calculators and we get 139.8 tools. And then our last step, our network is equal to our change in kinetic energy. So we can rewrite that as network is equal to one half M B F squared minus one half M V I S square. And we were told in the problem that the trolley started at rest. So V I is equal to zero. So this full term goes to zero. And now we have our network of 139 0.8 Jews is equal to one half mass is 15 kilograms multiplied by V F square. So we can rearrange that and we get V F is equal to two multiplied by 139.8 jewels divide divided by 15 kg. So let me take a third of that and we get 4.32 m per second. And so that is the answer to this problem. And that aligns with answer choice B. So B is the correct answer here. That's all we have for this one. We'll see you in the next video.

(b) Doug uses a 25 N horizontal force to push a 5.0 kg crate up a 2.0-m-high, 20° frictionless slope. What is the speed of the crate at the top of the slope?

Verified Solution

Key Concepts

Newton's Second Law of Motion

Work-Energy Principle

Kinematics