A hockey puck is given an initial speed of 5.3 m/s . If the coefficient of kinetic friction between the puck and the ice is 0.05, how far does the puck slide before coming to rest? Solve this problem using conservation of energy.
+1315 This is easy to do and it is very cool how it works.
Force here is what converts the kinetic energy to heat by friction and slows the puck
fu = coefficient of Kinetic friction
g = is the acceleration of gravity
Fu*g = a.
So friction force on the puck is the acceleration of gravity times the coefficient of kinetic friction and the acceleration force is made negative here because it is slowing down the puck. .
Puck friction to negative acceleration
fu=0.05
g = -9.81m/s^2
so
a = 0.05 * -9.81m/s^2 = -0.4905 m/s^2
For the moving puck
Velocity = v = u + (a*t) | u=initial velocity
So
Time = t = (v-u)/a
Distance = d = t *(v + u)/2
So
Distance d= ((v-u)/a) * ((v+u)/2)
This becomes d = (v-u)(v+u)/2a
Plugging in the numbers
d= (0 - 5.3)(0 + 5.3)/(2*-0.49.5) =28.63 meters
This is very slippery ice.
+1315 This is easy to do and it is very cool how it works.
Force here is what converts the kinetic energy to heat by friction and slows the puck
fu = coefficient of Kinetic friction
g = is the acceleration of gravity
Fu*g = a.
So friction force on the puck is the acceleration of gravity times the coefficient of kinetic friction and the acceleration force is made negative here because it is slowing down the puck. .
Puck friction to negative acceleration
fu=0.05
g = -9.81m/s^2
so
a = 0.05 * -9.81m/s^2 = -0.4905 m/s^2
For the moving puck
Velocity = v = u + (a*t) | u=initial velocity
So
Time = t = (v-u)/a
Distance = d = t *(v + u)/2
So
Distance d= ((v-u)/a) * ((v+u)/2)
This becomes d = (v-u)(v+u)/2a
Plugging in the numbers
d= (0 - 5.3)(0 + 5.3)/(2*-0.49.5) =28.63 meters
This is very slippery ice.
