Question

A 255 N force is pulling an 94.6-kg refrigerator across a horizontal surface. The force acts at an angle of 16.7 ° above the surface. The coefficient of kinetic friction is 0.173, and the refrigerator moves a distance of 5.88 m. Find (a) the work done by the pulling force, and (b) the work done by the kinetic frictional force.

Answer 1

A 255 N force is pulling a 94.6-kg refrigerator across a horizontal surface. The force acts at an angle of $16.7{}^{\circ}$ above the surface. The coefficient of kinetic friction is 0.173, and the refrigerator moves a distance of 5.88 m. Find

(a) the work done by the pulling force

(b) the work done by the kinetic frictional force.

$F_{fr}$ - friction force

$F$ - pulling force

a) Work can be expressed by the following equation:

$$W = F d \cos \theta$$

where $F$ is the force, $d$ is the displacement, and the angle $\theta$ is defined as the angle between the force and the displacement vector.

Therefore work of pulling force equals:

$$W _ {p} = 2 5 5 N * 5. 8 8 m * \cos 1 6. 7 = 1 4 3 6 J$$

Answer: 1436 J

b) Newton's first law of motion on y-axis:

$$F \sin (1 6. 7) + N = m g$$

Therefore, normal force equals:

$$N = m g - F \sin (1 6. 7)$$

Force of friction equals:

$$F_{fr} = \mu N = \mu (mg - F \sin (16.7))$$

work of friction force equals:

$$\begin{array}{l} W_f = F_{fr} d \cos 180 = -0.173 * (94.6 * 9.81 - 255 \sin 16.7) * 5.88 \\ = -869 J \end{array}$$

Answer: -869 J