A 200-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

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

ogorwyne ogorwyne · Answer 2 · 2022-03-28T14:50:33+02:00

The vertical height that the puck reach above its starting point is

16.847 m

Given

U = 16 m/s

m = 200 g

θ = 30°

µs = 0.40

µk = 0.30

the puck was moving at a initial velocity u due to friction and gravitational force the puck will stop at a point the height at this point is what we are looking for. from Newtons law of motion

[tex]v^{2} = u^{2} +2as[/tex]

v is final velocity which should be zero

u is initial velocity

a is negative acceleration

s is distance which is the required height

0 = [tex]u^{2} + 2as[/tex]

s = [tex]\frac{- (u^{2} )}{2a}[/tex]

to find the acceleration (deceleration) we resolve the forces as follows

sum of vertical forces is zero

N is Normal reaction

g is acceleration due to gravity = 10

N = m g cos θ

considering friction

vertical force Fv = N µk = m g cos θ µk

Sum of horizontal forces Fh = m a

m a = -(m g sin θ) - (m g cos θ) µk

a = - g (sin θ + cos θ µk)

a = - (10 (sin 30 + cos 30 * 0.3))

a = - 7.598 m/[tex]s^{2}[/tex]

solving for s gives

s = [tex]\frac{- (u^{2} )}{-2a}[/tex]

[tex]s = \frac{16^{2} } {2*7.598}[/tex]

[tex]s = 16.847 m[/tex]

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