Question

A crane lowers a stationary 200-kg crate 5-m to the ground. Right before hitting the ground, the crate is moving at 1 m/s. How much work is done by tension during this lowering process?

Answer 1

The work done by tension when lowering a 200-kg crate 5 meters is 100 joules, calculated by subtracting the kinetic energy before impact from the potential energy lost due to gravity.

Step-by-step explanation:

The work done by tension during the lowering process can be calculated using the work-energy principle. The work done is equal to the change in kinetic energy of the crate.

To calculate the change in kinetic energy, we need to find the initial kinetic energy and the final kinetic energy of the crate. The initial kinetic energy is zero since the crate is stationary, and the final kinetic energy can be calculated using the formula:

Final Kinetic Energy = 0.5 * mass * velocity^2

Substituting the given values, we get:

Final Kinetic Energy = 0.5 * 200 kg * (1 m/s)^2 = 100 J

So, the work done by tension during the lowering process is 100 J.