Question

A force of 25.0 N is applied 30° above the horizontal to a 5.0 kg box to move it across a smooth surface having coefficient of kinetic friction 0.2. Draw and label all the forces and calculate the acceleration of the box.

Answer 1

Answer:2.87

Explanation:

Given

mass of force=25 N

mass of box=5 kg

coefficient of kinetic friction $\mu _k=0.2$

Acceleration of box will be provided by its cos component and friction try to oppose it

Normal reaction$=mg-F\sin 30$

$N=5\times 9.8 -25\sin 30 =36.5 N$

friction force$(f_r)=\mu _k N=0.2\times 36.5=7.3 N$

net force in horizontal direction

$F\cos 30 -f_r=ma$

$25\cos 30-7.3=5\times a$

$a=2.87 m/s^2$