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

Step-by-step 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* 9.8 -25\sin 30 =36.5 N`

friction force
`(f_r)=\mu _k N=0.2* 36.5=7.3 N`

net force in horizontal direction

`F\cos 30 -f_r=ma`

`25\cos 30-7.3=5* a`

`a=2.87 m/s^2`