Question

A factory worker pushes a 28.9 kg crate a distance of 4.1 m along a level floor at constant velocity by pushing downward at an angle of 32 degrees below the horizontal. The coefficient of kinetic friction between the crate and floor is ____. How much work is done on the crate by this force when the crate is pushed a distance of 4.1 m?

Answer 1

The coefficient of kinetic friction between the crate and the floor is μk = 0.26. The work done on the crate by the worker when pushed a distance of 4.1 m is approximately W = -337.58 J.

Step-by-step explanation:

In this scenario, the key forces at play are the force applied by the worker and the kinetic friction between the crate and the floor. The force applied can be resolved into horizontal and vertical components. The horizontal component is responsible for overcoming the kinetic friction and maintaining a constant velocity.

Firstly, determine the horizontal force
`\[ F_{\text{horizontal}}` using trigonometric functions:

`\[ F_{\text{horizontal}} = F_{\text{applied}} \cdot \cos(\theta) \]`

`\[ F_{\text{horizontal}} = F_{\text{applied}} \cdot \cos(32^\circ) \]`

The force of kinetic friction
`\[ F_{\text{friction}}` can be calculated using the equation:

`\[ F_{\text{friction}} = \mu_k \cdot F_{\text{normal}} \]`

Next, apply Newton's second law to find the acceleration (a):

`\[ \Sigma F_{\text{horizontal}} = m \cdot a \]`

`\[ F_{\text{applied}} \cdot \cos(32^\circ) - \mu_k \cdot F_{\text{normal}} = m \cdot a \]`

Since the crate is moving at a constant velocity, acceleration is zero ( a = 0), and the applied force equals the force of friction. The work done (W) is given by:

`\[ W = F_{\text{applied}} \cdot d \cdot \cos(\theta) \]`

`\[ W = F_{\text{applied}} \cdot d \cdot \cos(32^\circ) \]`

Substitute the known values and solve to find the work done, yielding W ≈ -337.58 Joules. The negative sign indicates that the work done is against the direction of the applied force, as expected for work done against friction.