Question

a cable pulls a crate of mass 19.0 kg at a constant speed over a frictionless ramp at an angle 20.1° above the ground. if the total distance traveled is 5.40 m, find the work done by the cable on the crate.

Explanation:

Step1: Find the force applied by the cable

Since the crate is moving at a constant speed, the net force along the ramp is zero. The force component of the gravitational force along the ramp is \(mg\sin\theta\), and the force applied by the cable \(F\) is equal to this (because \(F - mg\sin\theta=0\)). Here, \(m = 19.0\space kg\), \(g = 9.8\space m/s^{2}\), \(\theta=20.1^{\circ}\). So \(F=mg\sin\theta\).

$$F=(19.0)(9.8)\sin(20.1^{\circ})$$

$$F = 19.0\times9.8\times0.344$$

$$F=64.1\space N$$

Step2: Calculate the work done

The formula for work done is \(W = Fd\), where \(d = 5.40\space m\) and \(F\) is the force along the direction of displacement.

$$W=(64.1)(5.40)$$

$$W = 346\space J$$

Answer:

\(346\space J\)