A 34.5 kg box initially at rest is pushed 5.80 m along a rough, horizontal floor

Question

A 34.5 kg box initially at rest is pushed 5.80 m along a rough, horizontal floor with a constant applied horizontal force of 135 N. If the coefficient of friction between box and floor is 0.300, find the following.

(a) the work done by the applied force
J

(b) the increase in internal energy in the box-floor system due to friction
J

(c) the work done by the normal force
J

(d) the work done by the gravitational force
J

(e) the change in kinetic energy of the box
J

(f) the final speed of the box
m/s

Explanation / Answer

Here,

a)

work done by applied force = force * distance

work done by applied force = 135 * 5.8

work done by applied force = 783 J

the work done by applied force is 783 J

b)

increase in internal energy = work done by frictional force

increase in internal energy = 0.3 * 34.5 * 9.8 * 5.80

increase in internal energy = 588.3 J

the increase in internal energy is 588.3 J

c)

as the noraml force is very perpendicular to the displacement

work done by normal force = 0 J

d)

as the gravitational force is perpendicular to displacemnt

work done by gravitational force = 0 J

e)

change in kinetic energy = net work done

change in kinetic energy = 783 - 588.3

change in kinetic energy = 194.7 J

the change in kinetic energy is 194.7 J

f)

let the final speed is v

0.5 * 34.5 * v^2 = 194.7

v = 3.36 m/s

the final speed of the box is 3.36 m/s

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