1) Consider a car that slides to a stop. If the initial speed doubles, I can see the kinetic energy at the beginning is 4 times greater, but why does this mean that your new stopping distance will be 4 times the old one? 2) How would you demonstrate, using a pen and a sheet of paper, that static friction can do positive work on a pen? [If I roll the pen on the paper, is that static friction and is that positive work? How can friction do positive work? Does kinetic friciton always do negative work?] 3) A toboggan is initially moving at a constant velocity along a snowy horizontal surface where friiction is negligible. When a pulling force is applied parallel to the ground over a certain distance, the kinetic energy increases by 47%. By what percentage woud the kinetic energy have changed if the pulling force had been at an angle of 38 degrees above the horizontal? [The answer is 16% but I can't figure out how to do it...] 4) Can the Work-Energy Theorem (W=change in kinetic energy=Ek2-Ek1 only be used when the force and displacement are pointing in the same direction? (0 degrees to each other) Because when we derive this equation, we used theta = 0 degrees. For example, would the above equation work in a question like this? (not every force is in exactly the same direction as the displacement...) "A 61-kg skier, coasting down a hill that is at 23 degrees to the horizontal, experiences a force of kinetic friction of 72N. The skier's speed is 3.5m/s near the top of the slope. Find the speed after the skier has travelled 62m downhill." Thanks for helping! Please Register or Log in to view the hidden image!

Energy is a very difficult thing for me to conceptualize. It seems to be a scalar quantity that is defined as the integral of momentum wrt velocity, but if anyone can help me find a way to think of it as a fundamental concept that would help a lot.

How is Kinetic Energy related to Force and Distance? How might you make a stationary pen move across a horizontal table, with no forces on the pen except gravity and static friction? The pen does not need to roll. No. Here's a thought for you: Friction always tries to equalise the velocity of the two objects in contact. It is not necessary that either is stationary. How is Kinetic Energy related to Force and Distance? What changes when the pulling force is applied at an angle instead of in the same direction as the motion? It can be used in all cases, as long as no Work is lost (as heat, deformation of the objects, etc). For example, gravity performs no work on a planet in a circular orbit, because it's kinetic energy is constant. This is true even though the force of gravity is continually accelerating the planet.

Hello, 1) Ek= 1/2 mv^2 It relates to speed...but how does it relate to distance? 2) But friction points in the opposite direction, and W=f * d cos theta d=magnitude of displacement d and Ff always have opposite direction, so cos 180 = -1 and W is negative. How can work done by friction possibly be positive? 3) W would be smaller if the force is applied at an angle to the horizontal. But is the answer 16% correct? I tried this question again but I got 37% or so... 4) Ek=kinetic energy Ep=gravitational potential energy Since W=delta Ek and W=delta Ep, is it correct to say that delta Ek = delta Ep?

dW = F*dr where F*dr is the dot product of the force and displacement vectors and the derivation of the work energy theorem has nothing to do with angle as far as I know Pete are you sure kinetic friction does not always do negative work? Given the above equation I can't think of any instances where the force of kinetic friction and the displacement point in the same direction. Remember, for kinetic friction to do POSITIVE work, it must be increasing the KE of the object. This is obviously impossible.

Kinetic friction can do positive work. Indeed, the work done on an object depends on the reference frame used to describe the motion. For example, consider a block that slides on a rough surface in a frame where the surface is at rest. The block begins with some initial kinetic energy, and friction slows the block until the relative velocity between the block and the surface is zero. The work done is negative since the kinetic energy decreased. Now imagine the same system in a frame where the block is initially at rest and the surface moves backwards. The block starts with no kinetic energy, but the friction force accelerates the block until it is moving with the surface. The work done is positive since the block increased its kinetic energy. There are two important things to take from this example. First, as Pete has emphasized, it is the relative velocity that friction tries to oppose. And second, the work done depends on the frame of reference you use.

Physics Monkey, I guess that works. But 'normal' people don't imagine the surface to be moving backwards while the block is at rest. Please Register or Log in to view the hidden image!

But, according to the people living on the block the opposite is true! They told me so. Please Register or Log in to view the hidden image!

Hi kingwinner, 1) If you double the initial speed then you quadruple the initial kinetic energy. The car still experiences the same constant force of friction trying to slow it down. The car stops when all the kinetic energy has been dissipated by friction. What is the work done by a constant force over a distance x? Answer this question and the answer to your original question will become obvious. 3) Assuming the 38 degree force has the same magnitude and is applied for the same distance, then the answer provided is wrong. What did you calculate the answer to be?

1) Work done by friction=Ff * d cos 180 = -Ff * d The initial kinetic energy quadruples, but why does that mean the stopping distance will also be quadrupled? 3) Are the initial kinetic energy the same (e.g. 100 J) for both the cases when the force is applied horizontally and at an angle of 38 degrees?

1) If you could originally stop the car in a distance D, how far would it take to stop a car with quadruple the initial kinetic energy? Remember, all the kinetic energy has to be dissipated, so just figure out how far the car has gone when all the kinetic is gone. You know how to do that. 2) I would assume the initial kinetic energies are the same, but I don't really know. Based on the way you worded the problem, it seems like that's a reasonable assumption. If the problem was asking for something else, I don't know what it is.