Question

Problem 2:

A 12-kg box is released from the top of an incline that is 5.0m long and makes an angle of 40° to the horizontal. A 60-N friction force impedes the motion of the box.

a) What will be the acceleration of the box?

b) How long will it take to reach the bottom of the incline?

c) What is the coefficient of friction between box and incline?

Answer 1

answer) here for balancing we have the eqn

Fnet=mgsin$\Theta$-Fr.........1)

from newtons law

Fnet=ma.........1)

so we have

a=mgsin$\Theta$-Fr/m=12*9,8*sin40-60/12=1.3 m/s²

so answer is 1.3 m/s²

b) from kinematics we have

s=ut+1/2at²

here u=0 ( since we start from rest)

so t=$\sqrt{}$2s/a=$\sqrt{}$2*5m/1.3=2.8 s

so answer is 2.8 s

c) we know the friction force is

Fr=${\mu }$Fn

Fn=mg*cos theta ( normal force)

${\mu }$=Fr/mgcos theta=60N/12*9.8*cos40=0.67

so answer is 0.67