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.
