A 35.0-N crate is pulled up a 4.0-m inclined plane by a worker at constant velocity.If the plane is inclined at an angle of 37° to the horizontal and there exists a constant frictional force of 25.0 N between the crate and the surface,what is the force applied by the worker?
A)zero
B)3.95 N
C)21.1 N
D)46.1 N
E)84.2 N

Question

Quizplus · Accepted Answer

To find the force applied by the worker, we need to consider both the component of the gravitational force acting down the incline and the frictional force. The gravitational force component acting down the incline can be calculated using $F_g = mg \sin(\theta)$, where $m$ is the mass of the crate, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $\theta$ is the angle of the incline. The mass of the crate is $35.0 \, \text{N} / 9.8 \, \text{m/s}^2 = 3.57 \, \text{kg}$. Thus, $F_g = 3.57 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times \sin(37^\circ) \approx 21.1 \, \text{N}$. The total force needed to pull the crate up at constant velocity equals the sum of this gravitational force component and the frictional force, which is $21.1 \, \text{N} + 25.0 \, \text{N} = 46.1 \, \text{N}$. Therefore, the worker must apply a force of 46.1 N to move the crate up the incline at constant velocity.

A 350-N Crate Is Pulled Up a 4