Vf^1=Vi^2 + 2ad but a= force/mass, so
you can solve for force.
coefficent offriction:
frictionforce=mu*mg solve for mu.
Hockey Puck A 110 g hockey puck sent sliding over ice is stopped in 13 m by the frictional force on it from the ice.
(a) If its initial speed is 5.0 m/s, what is the magnitude of the frictional force?
1 N
(b) What is the coefficient of friction between the puck and the ice?
2
3 answers
you have to give formulas not answers
v = u + a*t
so t = (v-u)/a
x = t (v+u)/2
substituting for t, x = (v+u)(v-u)/2a
put in known values:
13 = (5)(-5)/2a
a = -25/26 m/s^2
f = m*a
u(k) * f(normal) = m*a
u(k) *m*g = m*a
u(k) = a/g
u(k) = -25/26 /9.81 = 0.098
where u(k) = coefficient of friction
for magnitude of friction force:
u(k)*m*g = 0.098 * 0.110kg * 9.81 m/s^2 = 0.106N
so t = (v-u)/a
x = t (v+u)/2
substituting for t, x = (v+u)(v-u)/2a
put in known values:
13 = (5)(-5)/2a
a = -25/26 m/s^2
f = m*a
u(k) * f(normal) = m*a
u(k) *m*g = m*a
u(k) = a/g
u(k) = -25/26 /9.81 = 0.098
where u(k) = coefficient of friction
for magnitude of friction force:
u(k)*m*g = 0.098 * 0.110kg * 9.81 m/s^2 = 0.106N
1 answer