Question

The coefficient of kinetic friction between the inclined plane and the block is 0.7. The has a mass of 15 Kq. It is observed that when the block passes a point located 2 ong the plane, it is moving at a velocity of 1.6 m/sec.

a) What is the normal force of the block?

b) What is the friction force acting on the block?

c) What is the acceleration of the block?

d) What is the velocity of the block as it passes a point 5 m up on the plane?

Answer 1

coefficient of friction = tan

u = tan

= tan^-1 ( 0.7) = 35^o

a) normal force = component of weight perpendicular to the plane

= mg cos

= 15*9.8*cos 35

= 120.43 N

b) friction force = uN = 0.7*120.43 = 84.3 N

c) acceleration = (Ff - mg sin ) / m

= (84.3 - 15*9.8*sin 35) / 15

= -0.0011 m/s2

d) v^2 - u^2 = 2aS

v^2 - 1.6^2 = 2*-0.0011*5

v = 1.596 m/s