A1200-n crate rests on the floor. how much work is required to move it at constant speed (a) 5.0 m along the floor against a friction force of 230 n, and (b) 5.0 m vertically?

(a) The work required to move the crate = 7150 J

(b) The Work required to move the crate vertically = 6000 J

Explanation:

(a) Frictional Force: These is the force that tend to oppose the motion of  two bodies in contact.

Work required to move the crate = work done against friction + work required to move it through a distance of 5.0 m.

Wr = Wf + Wd

Where Wr = work required to move the crate, Wf = work done against friction, Wd = work required to move the crate through 5.0 m

Given: Wf = friction force × distance = 230×5 = 1150 J

and Wd = force × distance = 1200×5.0 =6000 J.

Substituting these values into equation 1

Wr = 6000 + 1150

Wr = 7150 J

Therefore the work required to move the crate = 7150 J

(b)

Work required to move the crate vertically = mgh

or

Work required to move the crate vertically =  Wh Equation 2

Where W = weight of the block, h = vertical height

Given:W = 1200 N, h= 5.0 m

Substituting these values into equation 2

Work required to move the crate vertically = 1200×5

Work required to move the crate vertically = 6000 J

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the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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