Physics, block down frictionless ramp?

Question

Hi,

I have a physics midterm today and this is one of the review problems:

A block of mass m starts from rest at height h. It slides down a frictionless incline, across a rough horizontal surface of length L, then up a frictionless incline. The coefficient of kinetic friction on the rough surface is Uk.

a) What is the block's speed at the bottom of the first incline?
b) How high does the block go on the second incline?

Give your answers in terms of m, h, L, Uk, and g.


Help would be greatly appreciated!


Thank you!

Answer 1

Since the first incline is frictionless, there is no energy loss in coming down. Hence KE at the foot of the incline = PE at the start. This gives the velocity at the foot of the first incline as
1/2 m*v^2 = m*g*h.
The frictional force on the horizontal surface = Uk*m*g
The work done in overcoming the fricional force =Uk*m*g*L
Thus the available energy at the foot of the second incline = m*g*h - Uk*m*g*L = m*g*(h - Uk*L)
If the height to which the block rises is h1,
m*g*h1 = m*g*(h - Uk*L). Hence
h1 = h - Uk*L

Answer 2

Let's think about this in terms of energy. At rest, the block has nothing but potential energy, expressed as m*g*h1, where h1 is the height of the first incline. Then the block slides down the frictionless ramp, converting all the potential energy to kinetic energy, or 0.5*m*v^2.

For problem a) we set m*g*h1 equal to 0.5*m*v^2 (since all the potential energy is converted to kinetic energy) and solve for v, the velocity after going down the first incline. So we get v^2=2*g*h1. Take the square root of (2*g*h1) to solve for v.

For the second part of the problem, the block rides across the flat, rough surface, which absorbs some energy. The amount of energy absorbed from that friction is given as m*g*Uk*L. Also, we know that the height the block goes on the second incline is dependent on the energy it has after sliding across that surface, when all the kinetic energy is converted back into potential energy.

The energy the block has before going up the second incline is equal to the energy it has at the bottom of the first, minus the energy lost due to friction. In mathematical terms, (m*g*h1) - (m*g*Uk*L) = m*g*h2. Solving for h2, we see that the terms m and g are common to both sides of the equation, so we can rewrite it to say h1 - (Uk*L) = h2. That's your answer!

Good luck on your midterm!!

Answer 3

a) The block starts with a potential energy m*g*h. This energy is converted to kinetic energy (1/2)*m*v^2 at the bottom. Solve for v.

b) While crossing the rough patch, friction Uk*m*g (N=m*g because surface is horizontal) does work Uk*m*g*L in slowing the block. The remaining kinetic energy at the beginning of the 2nd incline is original KE - work done by friction:
(1/2)*m*v^2 - Uk*m*g*L
Convert this kinetic energy back to potential energy and solve for the new height.

Answer 4

Let us consider the forces acting on the mass m. Let the angle of the incline be z. Then we have sin z=h/L, cos z=sqrt[L^2-h^2]/L.
The weight=mg acting down wards.
Component of weight along the incline=mg sin z
Normal reaction=R
Component of weight vertical to incline=mg cos z
Normal reaction R=m g cos z
Net force acting on the mass along the incline
=m g sin z-Uk.m g cos z
Net acceleration=g[sin z-Uk cos z]
We now apply the kinematic equation along the incline.
V^2-u^2=2gs
u=0
g=g[sin z-Uk cos z]
s=L
V^2=2*g[sin z-Uk cos z]*L
V=sqrt[2g(sin z-Uk cos z)*L]
sin z=h/L, cos z=sqrt[L^2-h^2]/L
This is the velocity at the base of the incline.
To know how how much it will travel on the other incline we need to know the angle of the inclined plane which is not specified.
If the angle is assumed to be z the same as the other then
height of travel=V^2/2g{ based on conservation of energy}
V=is the velocity at base
g=net acceleration=gsinZ[since there is no kinetic friction]
height=2g[sinz-Ukcos z]*L/2gsinz
=[1-Uk*cotz]*L
=[1-Uk*sqrt(L^2-h^2)/h]*L

Answer 5

The block, which does no longer rotate, arrives on the backside first. Of the solid ball and hollow sphere, the ball has the decrease 2nd of inertia, so it arrives 2nd. The hollow sphere arrives final.