Answer:
h = 8.588 m
Step-by-step explanation:
Given:
Mass of hockey puck: m = 200 g = 0.2 K g
Angle of incline: θ = 30º (with respect to horizontal)
Coefficient of static and kinetic friction between the puck and ramp:
μ s = 0.4 and μk = 0.3
initial speed:
v = 3.5 m /s
Let h be the vertical height reached by the puck, above the ground. This corresponds to a distance of
d = h *sin 30º = 2 *h
along the incline. As the puck is moving, only kinetic friction comes into play as it climbs up the incline. Let N be the normal reaction exerted by the metal ramp on the ball. Then,
N = m*g *cos θ = 0.2 *9.8 1*cos 30 º = 1.6991 N
Therefore, kinetic friction acting on the puck is:
F k = μ k *N = 0.3 *1.6974 = 0.50974 N
From work energy theorem, the change in kinetic energy should equal the work done by friction and gravity. Therefore,
0.5* 0.2 *16² = 0.50974 *2h + 0.2*9.81*h
⇒ h = 8.588 m