Question

A 46-kg box is being pushed a distance of 7.5 m across the floor by a force whose magnitude is 191 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.20. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force.Note: the four forces are the force P,the frictional force f, the normal forceN, and the force due to gravitymg.

Answer 1

Answer:

676.2 J

Explanation:

mass m=46 kg

displacement S= 7.5 m

coefficient of kinetic friction is 0.20.

The applied force does work;

$W_P= FScos\theta = 191\times7.5cos0= 1432.5$

The frictional force

$W_f=f_kScos180$= $-\mu_k F_N S$

Now F_N= mg = 46*9.8

therefore,

$W_f=-\0.2\times46\times9.81\times7.5$= -676.2 J

The normal force and gravity do NO WORK,because they act at 90 degrees to displacement.

the work done on the box by each of the four forces that act on the box= 676.2 J