A 40kg crate is pulled 40m along a horizontal floor by a constant force exerted

Question

A 40kg crate is pulled 40m along a horizontal floor by a constant force exerted by a person FP = 100N, which acts at a 370 angle as in the picture below.

If the velocity is constant, determine the work done by

the force of the person pulling.

The gravitational force exerted by earth.

The normal force.

The frictional force

Calculate the total work. Show your work.

The pulling force of the person is increased to 125 N. The coefficient of kinetic friction is µk = 0.20 Find the acceleration of the box.

Find the work done by each force below when the force of the person is at 125 N. The box is pulled a distance of 2.5 m.

Gravitational force exerted by earth

Normal force

Frictional force

Force of the person

Calculate the total work done

Explanation / Answer

mass of crate= 40kg

distance = 40m

Force of Person = 100N,

angle is = 370

The work done by the gravitational (Wg) and normal forces (WN) are perpendicular to the displacement (x), and so

Wg = OJ

WN = OJ

This is because the q =90o Cos(90o)=0

The work done by Fp is

Wp= Fpxcos(theta) = (100N)(40m) cos 37o = 3194.54 J

The work done by the friction force is

Wf = Ffrxcos (180) = (50N)(40m) (-1) = -2000J

The angle is 180o because x and F for point in opposite directions. Since cos 180o = -1, we see that the force of friction does negative work on the crate. Finally, the net work is: Wnet = Wg + WN + Wp + Wf = 0 + 0 + 3200 - 2000 = 1200J

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