Answer:
a) 11.8 m/s
b) 0.65 m
c) 7.0 m
Explanation:
Energy is conserved. The block starts with gravitational potential energy, which is converted to kinetic energy and work done by friction. The friction force is equal to the coefficient of friction times the normal force, which can be found by drawing a free body diagram. After the ramp, the block's kinetic energy is converted to elastic potential energy, and then back to gravitational potential energy and work done by friction.
a) Drawing a free body diagram, there are 3 forces on the block while it is on the ramp: weight force mg pulling down, normal force N pushing up perpendicular to the incline, and friction force Nμ pushing up parallel to the incline. Sum of forces in the perpendicular direction:
∑F = ma
N − mg cos θ = 0
N = mg cos θ
Therefore, the friction force is Nμ = mgμ cos θ.
Energy is conserved. Initial potential energy equals final kinetic energy plus work done by friction.
PE = KE + W
mgh = ½ mv² + Fd
mg (L sin θ) = ½ mv² + mgμL cos θ
gL sin θ = ½ v² + gμL cos θ
½ v² = gL sin θ − gμL cos θ
v² = 2gL (sin θ − μ cos θ)
Plugging in values:
v² = 2 (9.8) (10) (sin 60 − 0.3 cos 60)
v = 11.8 m/s
b) Energy is conserved. Initial kinetic energy equals final elastic energy.
KE = EE
½ mv² = ½ kx²
mv² = kx²
x = v √(m / k)
Plugging in values:
x = 11.8 √(1.5 / 500)
x = 0.65 m
c) Energy is conserved. Initial kinetic energy equals final potential energy plus work done by friction.
KE = PE + W
½ mv² = mgh + Fd
½ mv² = mgd sin θ + mgμd cos θ
v² = 2gd (sin θ + μ cos θ)
d = v² / [2g (sin θ + μ cos θ)]
Plugging in values:
d = (11.8)² / [2 (9.8) (sin 60 + 0.3 cos θ)]
d = 7.0 m
