A 100 lb block of ice slides down an incline 5.0 ft long and 3.0 ft high. A worker pushes up on the ice parallel to the incline so that it slides down at constant speed. The coefficient of friction between the ice and the incline is 0.10.
Find (a) the force exerted by the worker, (b) the work done by the worker on the block, (c) the work done by gravity on the block, (d) the work done by the surface of the incline on the block, (e) the work done by the resultant force on the block, and (f) the change in kinetic energy of the block.
I solved a and b, and I get them, but I'm stuck conceptually on the last few parts. On c, my biggest issue is that I think I'm confusing this situation to one where a block attached to a string is lowered (i.e. negative work is done on the block.)
Does (d) even matter since it's not the incline pushing the block down it? Or does friction play a part here?
For (e), if Fnet is 0 (since all the forces have to be 0 for there to be constant speed) does that mean this question is irrelevant?
...I can figure (f) out on my own.
I don't want answers, just...some conceptual help on this one.
I can definitely provide you with some conceptual help on this problem.
Let's start with part (c) - the work done by gravity on the block. In this scenario, the block is sliding down the incline at a constant speed, which means there is no acceleration. Therefore, the net force on the block must be zero.
Since the incline is inclined at an angle, gravity can be resolved into two components: one parallel to the incline, called the gravitational force component along the incline (mg*sinθ), and one perpendicular to the incline, called the gravitational force component perpendicular to the incline (mg*cosθ). Here, θ is the angle of the incline.
Since the block is not moving vertically, the gravitational force component perpendicular to the incline does not perform any work on the block. The work done by gravity is only the product of the gravitational force component along the incline and the displacement along the incline.
Now, let's move on to part (d) - the work done by the surface of the incline on the block. Since friction is involved, it does play a part here. Friction is the opposing force to the motion of the block, parallel to the incline. The work done by friction can be calculated as the product of the force of friction and the displacement along the incline.
In this case, the force of friction is the product of the coefficient of friction and the perpendicular component of the weight of the block (mg*cosθ). The displacement along the incline is the same as the length of the incline, given as 5.0 ft.
Finally, let's address part (e) - the work done by the resultant force on the block. Since the block is sliding down the incline at a constant speed, the net force on the block must be zero. This means that the work done by the resultant force is also zero. The individual forces (worker's force, gravity, and friction) all contribute to cancel each other out, resulting in no change in kinetic energy.
I hope this explanation provides the conceptual clarity you were looking for.