What is the total work done on the crate?
Understand the Problem
The question is asking for the total work done on a crate being pushed over a certain distance, considering the forces involved, such as friction. To solve it, we need to calculate the force of friction and apply the work formula (work = force × distance).
Answer
The total work done on the crate is approximately $34.73 \, \text{J}$.
Answer for screen readers
The total work done on the crate is approximately $W \approx 34.73 , \text{J}$.
Steps to Solve
- Calculate the Normal Force
Since the crate is on a level surface, the normal force ($F_N$) is equal to the weight of the crate. The weight can be calculated using the formula:
$$ F_N = m \cdot g $$
where:
- $m = 28.9 , \text{kg}$ (mass of the crate)
- $g = 9.81 , \text{m/s}^2$ (acceleration due to gravity)
- Determine the Force of Friction
Now, we can find the force of friction ($F_f$) using the equation:
$$ F_f = \mu_k \cdot F_N $$
where:
- $\mu_k = 0.26$ (coefficient of kinetic friction)
- Plug in the Normal Force
First, calculate the normal force:
$$ F_N = 28.9 \cdot 9.81 $$
Then, substitute into the friction formula:
$$ F_f = 0.26 \cdot F_N $$
- Calculate the Work Done Against Friction
Work done ($W$) against friction can be computed using the formula:
$$ W = F_f \cdot d $$
where:
- $d = 4.85 , \text{m}$ (distance the crate is pushed)
- Combine the Formulas to Find Work Done
From steps 2 & 4, the total work done can be expressed as:
$$ W = 0.26 \cdot (m \cdot g) \cdot d $$
- Final Calculation
Now substitute the values into the final equation:
$$ W = 0.26 \cdot (28.9 \cdot 9.81) \cdot 4.85 $$
The total work done on the crate is approximately $W \approx 34.73 , \text{J}$.
More Information
This answer considers both the mass of the crate and the kinetic friction, which acts against the crate's movement. The calculation highlights how friction affects the work done by the worker in pushing the crate.
Tips
- Forgetting to account for the direction of forces when calculating work.
- Not using the correct value for gravitational acceleration.
- Miscalculating the normal force.
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