Question

A block of mass m is set sliding up along a frictionless ramp inclined at angle θ relative to horizontal. The angle of the ramp can be varied, but the initial velocity (v) of the block does not vary. The block reaches a height h above its initial level where it momentarily stops. When referring to a change of a variable, it means running the experiment multiple times, each time with a different value for the variable in question.

What happens to the height h as the angle θ is increased?

What happens to the height h if θ and v are kept the same, but the mass m of the block increases?

If we keep m and v the same, but now we add a small amount of kinetic friction to the ramp surface, how does the height h change as θ increases?

Answer 1

Answer:

1) h is constant

2) h is constant

3) h increases

Explanation:

"What happens to the height h as the angle θ is increased?"

The ramp is frictionless, so the initial kinetic energy is converted to the final gravitational potential energy.

½ mv² = mgh

½ v² = gh

h = v² / (2g)

Since v is constant, h is also constant.

"What happens to the height h if θ and v are kept the same, but the mass m of the block increases?"

As found above, h does not depend on the mass m, so h would remain constant.

"If we keep m and v the same, but now we add a small amount of kinetic friction to the ramp surface, how does the height h change as θ increases?"

This time, some of the kinetic energy is converted into work done by friction.

½ mv² = Fd + mgh

½ mv² = Fh / sin θ + mgh

The friction force can be found by drawing a free body diagram of the block.  There are three forces acting on the block.  Normal force perpendicular to the ramp, weight force pulling down, and friction force pushing down the ramp.

Sum of the forces perpendicular to the ramp:

∑F = ma

N − mg cos θ = 0

N = mg cos θ

Friction is normal force times coefficient of friction:

F = Nμ

F = mgμ cos θ

Substituting:

½ mv² = (mgμ cos θ) h / sin θ + mgh

½ mv² = mgμ h / tan θ + mgh

½ v² = gμ h / tan θ + gh

½ v² = (gμ / tan θ + g) h

As θ increases, tan θ increases.  That makes gμ / tan θ + g decrease.  Since v is constant, h must increase.

Answer 2

Answer:

Explanation:

h is the height h above its initial level where the block momentarily stops - at that point, all its initial kinetic energy is converted to its potential energy.

The block's initial velocity (v) does not vary as the angle of the ramp varies.

By conservation of energy, the PE remains the same as as the angle θ is increased. So the height h remains the same.

At height h,  the block's initial kinetic energy is converted to its potential energy. So 1/2 * mass, m * v^2 =  mass, m * g * h

The mass, m, cancels each other out. So the height h remains the same as m increases.

If m and v are kept the same, a small amount of kinetic friction to the ramp surface will require energy to overcome its friction force as the block travels.

Energy loss = friction force * distance traveled

As θ increases, distance traveled decreases. Energy loss to friction decreases as well.

At height h,  the block's initial kinetic energy = potential energy + energy loss. As θ increases, energy loss decreases, potential energy increases. So h increases as well.