Question

A box slides down a ramp with inclination angle 40? with a distance function of:

s(t) = ( 63 cm/s ) t + ( 252 cm/s² ) t²

What is the coefficient of kinetic friction of the system?

Answer 1

gravitational force down the inclline Fg = m*g*sintheta

normal force N = m*g*costheta

frictional force fk = uk*N = uk*m*g*costheta

net force Fnet = Fg - fk

net force acting

Fnet = m*g*sintheta - uk*m*g*costheta

Fnet = m*a

m*a = m*g*sintheta - uk*m*g*costheta

acceleration a = g*(sintheta - uk*costheta)

distance s = vo*t + (1/2)*a*t^2

given s(t) = (63 cm/s)t + (252 cm/s^2 )t^2

therefore

(1/2)*a = 252

(1/2)*g*(sintheta - uk*costheta) = 252

g = 980 cm/s^2

(1/2)*980*(sin40 - uk*cos40) = 252

coefficient of kinetic friction uk = 0.168 <<<-----------ANSWER

DONE please check the answer. any doubts post in comment box