Question

What is the coefficient of kinetic friction between a 10-kg box and the surface as it slides at a constant velocity down a ramp which makes a 38 degree angle with the horizontal?

Answer 1

given
m = 10 kg
theta = 38 degrees

let mue_k is the coefficient of kinetic friction.

Normal force acting on the block, N = m*g*cos(theta)

As the block sliding with consatnt speed, net force acting on it must be zero.

so, net force along inclinded path = 0

m*g*sin(theta) - fk = 0

m*g*sin(theta) - mue_k*N = 0

m*g*sin(theta) - mue_k*m*g*cos(theta) = 0

m*g*sin(theta) = mue_k*m*g*cos(theta)

==> mue_k = sin(theta)/cos(theta)

= tan(theta)

= tan(38)

= 0.781 <<<<<<<<<<-------------------------Answer