A 57-kg box is being pushed a distance of 7.5 m across the floor by a force whose magnitude is 178 N. The force is parallel to the displacement of the box. The coefficient of kinetic friction is 0.19. 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.

Work by normal force is 1335 J

Work by frictional force is -796.8 J

Explanation:

Work done due to applied force is given by the product of magnitude of force and distance pushed.

Work done=Fd

Substituting 178 N for F and 7.5 m for d then

Work done by normal force=178*7.5=1335 J

Work done due to frictional force will be product of normal force, distance and coefficient of kinetic friction.

Work due to frictional force is -0.19*57*9.81*7.5=-796.81725 J

The normal and gravity forces do not work since they are at 90 degrees.

in order to follow the instructions and complete the statements, i need to use data from table c of your student guide. can you think of a way to let me see your student guide for a while ? i can't answer the question without the data in table c.