A 15-kg block is dragged over a rough, horizontal surface by an applied force of [01] N acting at 20.0 degrees above the horizontal. The block is displaced 5.00 m, and the coefficient of kinetic friction is 0.300. Find the work done by a. the applied force (J). b. the normal force (J). c. the force of gravity (J). d. the force of friction (J).

(a). The work done by the applied force is 328.89 J.

(b). The work done by the normal force is zero.

(c). The work done by the force of gravity is zero.

(d). The work done by the force of friction is 184.5 J.

Explanation:

Given that,

Force = 70 N

Mass of block = 15 kg

Angle = 20°

Displaced = 5.00 m

Coefficient of kinetic friction = 0.300

(a). We need to calculate the work done by the applied force

Using formula of work done

Put the value into the formula

(b). We need to calculate the work done by the normal force

The normal force is perpendicular to the displacement of the motion,

So it does not work.

(c). We need to calculate the work done by the force of gravity

The gravitational force is perpendicular to the displacement,

Or the displacement in the vertical direction is zero.so the work done is zero.

(d). We need to calculate the normal force

Using balance equation

Put the value into the formula

We need to calculate the force of friction

Using formula of friction

Put the value into the formula

We need to calculate the work done by the force of friction

Using formula of work done

Put the value into the formula

Hence, (a). The work done by the applied force is 328.89 J.

(b). The work done by the normal force is zero.

(c). The work done by the force of gravity is zero.

(d). The work done by the force of friction is 184.5 J.

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .