Answer:
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
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]
Read more on frictional force here: https://brainly.com/question/4618599
