KE= energy in friction
1/2 m v^2=mu*mg*distance
solve for distance.
a hockey puck is given the initial speed of 10 m/s. If the coefficient of kinetic friction between the ice and puck is 0.10, how far will the puck slide before stopping?
3 answers
What you wanna do is snort a line, smoke a J and look beyond what you see
f = ma
u(k) * f(normal) = m*a
u(k) *m*g=m*a
u(k) *g =a
a = 0.10 * 9.81m/s^2 = 0.981 m/s^2
this is defined to be a negative acceleration
v = u + a*t
so t = (v-u)/a
and x = t * (v+u)/2
so together x = (v+u)(v-u)/2a
where x = distance, v = final speed, u = initial speed, t = time, a = acceleration
substitute in our known values:
x = (0+10)(0-10)/(2*-0.981)
x = 50/0.981 = 50.97m
u(k) * f(normal) = m*a
u(k) *m*g=m*a
u(k) *g =a
a = 0.10 * 9.81m/s^2 = 0.981 m/s^2
this is defined to be a negative acceleration
v = u + a*t
so t = (v-u)/a
and x = t * (v+u)/2
so together x = (v+u)(v-u)/2a
where x = distance, v = final speed, u = initial speed, t = time, a = acceleration
substitute in our known values:
x = (0+10)(0-10)/(2*-0.981)
x = 50/0.981 = 50.97m
7 answers