Question

A 10 kg block is pushed across a rough surface by a 25N force the block moves at a constant velocity when this force is applied.
A) determine the magnitude of the friction force acting on the block
B) What is the coefficient of friction between the surface and the block

Answer 1

Fb = 10kg * 9.8N/kg = 98N.

Fb =98N @ 0 deg.

Fp = 98sin(0) = 0 = Force parallel to plane(hor).

Fv=98cos(0) = 98N.=Force perpendicular to the plane.

a. Fn = Fap - Fp - Ff = 0,
25 - 0 - Ff = 0,
Ff = 25N = Force of friction.

b. Ff = uFv = 25,
u*98 = 25,
u = 0.26 = Coefficient of friction.