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$\degree$ 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 the force P is doing work is given by the dot product of the force and velocity vectors. At t = 2.0 s, the velocity of the block is:$$v = \frac{P}{m}t = \frac{2.0 	ext{ N}}{3.0 	ext{ kg}}(2.0 	ext{ s}) = 1.33 	ext{ m/s}$$The force vector is:$$F = P\cos	heta \hat{i} + P\sin	heta \hat{j} = (2.0 	ext{ N})\cos 22^\circ \hat{i} + (2.0 	ext{ N})\sin 22^\circ \hat{j}$$The dot product of the force and velocity vectors is:$$P \cdot v = F \cdot v = (2.0 	ext{ N})\cos 22^\circ (1.33 	ext{ m/s}) + (2.0 	ext{ N})\sin 22^\circ (0) = 2.3 	ext{ W}$$Therefore, the rate at which the force P is doing work at t = 2.0 s is 2.3 W.

A 30-Kg Block Is on a Frictionless Horizontal Surface °\degree° Above the Horizontal Is Applied to the Block

A 30-Kg Block Is on a Frictionless Horizontal Surface $\degree$ Above the Horizontal Is Applied to the Block