Across a rough, horizontal surface. The chair's mass is 18.8 kg. The force you exert on the chair is 165 N directed 26 degrees below the horizontal. While you slide the chair a distance of 6.00 m , the chair's speed changes from 1.30 m/s to 2.50 m/s . Find the work done by friction on the chair.

-847.2J

Explanation:

First find the acceleration from v^2= u^2 + 2as

v= 2.5 m/s

u= 1.3 m/s

a???

s=6.00

a= v^2-u^2/2s

a= (2.5)^2-(1.3)^2/2× 6

a= 0.38ms^-2

From Newtons second law:

(Force applied cos Θ) - (Frictional force) = ma

Frictional force = ma- (Force applied cos Θ)

Frictional force= (18.8×0.38) - (165 cos 26°)

Frictional force= 7.144- 148.3= -141.2N

Therefore,

Work done by friction = Frictional force × distance covered

= -141.2N × 6= -847.2J

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