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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weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards