A200-g hockey puck is launched at an initial speed of 16 m/s 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 µs = 0.40 and µk = 0.30, respectively. what vertical height does the puck reach above its starting point

h = 8.588 m

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

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the answer is gallery.