A 3.0-kg block is on a frictionless horizontal surface. The block is at rest when, at t = 0, a force (magnitude P = 2.0 N) acting at an angle of 22° above the horizontal is applied to the block. At what rate is the force P doing work at t = 2.0 s?
A) 2.3 W
B) 2.0 W
C) 1.4 W
D) 1.7 W
E) 1.2 W

Question

Quizplus · Accepted Answer

The rate at which force P is doing work on the block is given by the power, which is the dot product of the force vector and the velocity vector of the block. Since the force is applied at an angle, only the horizontal component of the force contributes to the work done on the block. The horizontal component of the force is $P \cos(\theta) = 2.0 \, \text{N} \cdot \cos(22^\circ)$. The acceleration of the block is given by $a = \frac{F}{m}$, where $F$ is the horizontal component of the applied force and $m$ is the mass of the block. Thus, $a = \frac{2.0 \cos(22^\circ)}{3.0}$. The velocity of the block at $t = 2.0 \, \text{s}$ is $v = at = \frac{2.0 \cos(22^\circ)}{3.0} \cdot 2.0$. The power, or the rate at which work is done, is $P \cos(\theta) \cdot v$, which, when calculated with the given values, yields approximately 2.3 W.

A 30-Kg Block Is on a Frictionless Horizontal Surface