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# Physics Work Problem?

A 15.6 kg block is dragged over a rough, horizontal surface by a constant force of 77.5N acting at an angle of 27.4 degrees above the horizontal. The block is displaced at 54.2m, and the coefficient of friction is 0.155. The acceleration of gravity = 9.81.

What is the magnitude of the workd done by the force of friction. Answer in joules.

### 3 Answers

- MadhukarLv 71 decade agoFavorite Answer
Frictional force

= Normal force x coefficient of friction

= [(15.6) * (9.81) - 77.5 sin 27.4°] * (0.155)

= 18.19 N

Work done by frictional force

= Frictional force

= 18.19 * 54.2

= 986 J.

- 1 decade ago
W=Fnet x d where Fnet is the NET force and d is your displacement.

Fnet is the sum of the forces. There are four forces at work, the 77.5 N, say Fc, the force of friction, f, the force of gravity, Fg and the normal force, N.

First we look at the forces in the "Y" or up/down direction. Since the object is not moving up or down, the new forces in that direction sum to zero. The only forces acting in that direction are N, Fg and the Y component of the applied force, Fc. Fg opposes the other forces so, (Fc x sin(theta)) + N - Fg = 0. Therefore, N = Fg - (Fc x sin(theta)) where theta = 27.4

Now we look at the forces in the 'X' direction. The forces in this direction are f and the X component of Fc. The friction opposes the force applied so (Fc x cos(theta)) - f = Fnet. We know that f = mu x N, where mu is the coefficient of friction. Therefore, Fnet = (Fc x cos(theta)) - (mu x (Fg - (Fc x sin(theta)))). Fg = m x g or 15.6 x 9.81

- jgouldenLv 71 decade ago
The forces on the block are:

The normal force N, straight up

The friction force μN, straight back

Gravity mg, straight down

The pulling force P, with components

P cos 27.4 forward

P sin 27.4 up

The net force in the vertical dimension is zero, so

N + P sin 27.4 - mg = 0

Solve for n>

Then the motion in the horizontal dimension is

F = ma = P cos 27.4 - μN

Solve for F. Then work is force times distance.