Question012 (part 1 of 2) $\mathbf{1 0 . 0}$ points A 17.2 kg block is dragged over a rough, horizontal surface by a constant force of 89.6 N acting at an angle of $27.8^{\circ}$ above the horizontal. The block is displaced 33.7 m , and the coefficient of kinetic friction is 0.234 . $$\mu=0.234$$
Find the work done by the 89.6 N force. The acceleration of gravity is $9.8 \mathrm{~m} / \mathrm{s}^{2}$.
Answer in units of J. 013 (part 2 of 2) 10.0 points Find the magnitude of the work done by the force of friction.
Answer in units of J.

Studdy Solution
Calculate the work done by the force of friction using the formula:
$$W_f = f_k \cdot d \cdot \cos(180^\circ)$$
Since the friction force acts opposite to the direction of displacement, the angle is $180^\circ$, and $\cos(180^\circ) = -1$.
$$W_f = 29.64 \times 33.7 \times (-1)$$
$$W_f \approx -998.5 \, \text{J}$$
The work done by the 89.6 N force is approximately $2670.4 \, \text{J}$, and the work done by the force of friction is approximately $-998.5 \, \text{J}$.