How much work is done on the crate by this force? Express your answer using two significant figures.

Steps to Solve

1. Identify the forces acting on the crate

The only horizontal force acting on the crate is the applied force minus the force of friction. The crate moves at constant velocity, meaning the net force is zero.

2. Calculate the force of friction

The force of friction can be calculated using the formula:

$$ f_{\text{friction}} = \mu_k \cdot N $$

where:

• $\mu_k = 0.26$ (coefficient of kinetic friction)
• $N = m \cdot g$ (normal force, which equals the weight of the crate)

Here, the mass $m = 28.9 , \text{kg}$ and gravitational acceleration $g \approx 9.81 , \text{m/s}^2$.

Calculating the normal force:

$$ N = 28.9 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 283.5 , \text{N} $$

Now calculate the friction force:

$$ f_{\text{friction}} = 0.26 \cdot 283.5 , \text{N} = 73.8 , \text{N} $$

3. Calculate the work done by the applied force

Since the work done is equal to the force multiplied by the distance,

$$ W = F \cdot d $$

Here, $F$ (the applied force) is equal to the force of friction (since the crate moves at constant velocity), and $d = 4.85 , \text{m}$.

So,

$$ W = 73.8 , \text{N} \cdot 4.85 , \text{m} $$

Calculating the work done:

$$ W = 357.8 , \text{J} $$

4. Express the answer using two significant figures

Rounding to two significant figures, we get:

$$ W \approx 360 , \text{J} $$