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Slav-nsk [51]
2 years ago
14

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 k

inetic friction 0.2. Draw and label all the forces and calculate the acceleration of the box.

Physics
1 answer:
nordsb [41]2 years ago
8 0

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

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