Questions: A worker pushes a 1.50 x 10^3 N crate with a force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220. How much work is done by the worker on the crate?

A worker pushes a 1.50 x 10^3 N crate with a force of 345 N a distance of 24.0 m. Assume the coefficient of kinetic friction between the crate and the floor is 0.220. How much work is done by the worker on the crate?

Transcript text: 4. A worker pushes a $1.50 \times 10^{3} \mathrm{~N}$ crate with a force of 345 N a distance of 24.0 m . Assume the coefficient of kinetic friction between the crate and the floor is 0.220 . a. How much work is done by the worker on the crate?

Solution

Solution Steps

Step 1: Calculate the Work Done by the Worker

The work done by the worker on the crate can be calculated using the formula: \[ W = F \cdot d \cdot \cos(\theta) \] where:

$ W $ is the work done,
$ F $ is the force applied by the worker (345 N),
$ d $ is the distance over which the force is applied (24.0 m),
$ \theta $ is the angle between the force and the direction of motion (0 degrees, since the force is applied horizontally).

Since $ \cos(0^\circ) = 1 $: \[ W = 345 \, \text{N} \times 24.0 \, \text{m} \times 1 \]

Step 2: Perform the Calculation

\[ W = 345 \, \text{N} \times 24.0 \, \text{m} \] \[ W = 8280 \, \text{J} \]

The work done by the worker on the crate is $ 8280 \, \text{J} $.