Question

A box initially at rest is pushed along a rough, horizontal floor with a constant applied horizontal force of.The coefficient of friction between box and floor is . Find (a) the work done by the applied force, (b) the increase in internal energy in the box–floor system as a result of friction, (c) the work done by the normal force, (d) the work done by the gravitational force, (e) the change in kinetic energy of the box, and (f) the final speed of the box.

Short answer

(a) The work done by the applied force is 650J.

Step-by-step solution

Step-by-Step Solution Step 1: Given information

The mass of the box is m= 40 kg

The distance moved by the box is x=5.00 m

The horizontal pushing force applied to the box is, F=130N

The coefficient of friction between box and floor is, ${\mu }_k=0.300$

Determine the work done by force

If a body moves to a distance when a force is applied on it, then the internal energy of the body changes and a work is done on it.

The total change in the energy of the body by applying a force is balanced by the work done on the body by applied force

Step 3(a): Work done by the applied force

The free-body diagram of the box is given by,

Here, is the normal reaction force, and is the frictional force acting on the box.

The formula for the work done by the applied force on the box is given by,

w =Fx

$w=130\times 5\times \frac{1J}{1{\displaystyle \raisebox{1ex}{$kg.m^2$}\!\left/ \!\raisebox{-1ex}{$s^2$}\right.}}$

w=650J

Hence, the work done by the applied force on the box is 650J