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

Steps to Solve

1. Identify the values given We know:

• Mass of the crate, ( m = 28.9 , \text{kg} )
• Distance pushed, ( d = 4.85 , \text{m} )
• Coefficient of kinetic friction, ( \mu_k = 0.26 )

2. Calculate the force of friction The force of friction ( f_{\text{friction}} ) can be calculated using the equation: $$ f_{\text{friction}} = \mu_k \cdot N $$ Where ( N ) (the normal force) is equal to the weight of the crate, ( N = m \cdot g ), and ( g ) is the acceleration due to gravity ( g \approx 9.81 , \text{m/s}^2 ).

Calculating ( N ): $$ N = 28.9 , \text{kg} \cdot 9.81 , \text{m/s}^2 \approx 283 , \text{N} $$

Then substitute ( N ) to find ( f_{\text{friction}} ): $$ f_{\text{friction}} = 0.26 \cdot 283 \approx 73.58 , \text{N} $$

3. Calculate the work done by friction The work done ( W ) by the force of friction while the crate is pushed a distance ( d ) is given by: $$ W = f_{\text{friction}} \cdot d $$ Substituting the known values: $$ W = 73.58 , \text{N} \cdot 4.85 , \text{m} \approx 356.45 , \text{J} $$

4. Determine the sign of the work done Since friction opposes the motion, the work done by friction is negative: $$ W_{\text{friction}} = -356.45 , \text{J} $$

5. Round to two significant figures When rounding the work done to two significant figures, we get: $$ W_{\text{friction}} \approx -360 , \text{J} $$