A crate follows a semicircular path of radius 1.0 m. The tangential component of the net force acting on a crate is 10 N, constant in magnitude. The net work done on the crate is
A) 0 J.
B) 3.1 J.
C) 20 J.
D) 31 J.
E) 63 J.

Question

Quizplus · Accepted Answer

The net work done on the crate can be calculated using the formula for work, which is $W = F \cdot d \cdot \cos(\theta)$, where $F$ is the force applied, $d$ is the distance over which the force is applied, and $\theta$ is the angle between the force and the direction of motion. Since the force is tangential to the path, $\theta = 0^\circ$, making $\cos(\theta) = 1$. The distance $d$ is the length of the semicircular path, which can be calculated as half the circumference of a circle with radius $r = 1.0$ m, so $d = \pi \cdot r = \pi \cdot 1.0 \approx 3.14$ m. Thus, the work done is $W = 10 \, \text{N} \cdot 3.14 \, \text{m} = 31.4 \, \text{J}$, which rounds to $31 \, \text{J}$, making the correct answer D.

A Crate Follows a Semicircular Path of Radius 1