Question

A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients
of static and kinetic friction between the hockey puck and the metal ramp are #5 = 0.40 and pk =

0.30, respectively. The puck's initial speed is 14.9 m/s. What speed does it have when it slides back
down to its starting point?

Answer 1

Step 1: find the max vertical height achieved by puck in upward motion:

Using force balance along the incline:

F_net = W*sin + Fk

-m*a = m*g*sin + *m*g*cos

-a = g*sin + *g*cos

-a = 9.81*sin 30 deg + 0.30*9.81*cos 30 deg

a = -7.454 m/s^2 = acceleration of block (-ve since in opposite direction of motion)

U = Initial speed of puck 14.9 m/s

V = final speed after traveling max distance, d = 0 m/s

Using 3rd kinematic equation:

V^2 = U^2 + 2*a*d

d = (V^2 - U^2)/(2*a)

d = (0^2 - 14.9^2)/(2*(-7.454))

d = 14.89 m = distance traveled along the incline

Step 2: after traveling above distance puck stops and reverses it's direction, So now again using force balance

F_net = W*sin - Fk

m*a1 = m*g*sin - *m*g*cos

a1 = g*sin - *g*cos

a1 = 9.81*sin 30 deg - 0.30*9.81*cos 30 deg

a1 = 2.356 m/s^2 = acceleration of puck when traveling downward

U1 = Initial speed of puck when it reverses direction = 0 m/s

d1 = distance traveled to reach at starting point = 14.89 m

So, again using 3rd kinematic equation

V1^2 = U1^2 + 2*a1*d

V1 = sqrt (0^2 + 2*2.356*14.89)

V1 = final speed at starting point = 8.38 m/s

Let me know if you've any query.