How much work is done by the normal force?
Understand the Problem
The question is asking for the amount of work done by the normal force in a scenario where a crate is pushed horizontally along a level floor. Given the parameters such as mass, distance, and coefficient of kinetic friction, we need to determine how the normal force contributes to the work done in this situation.
Answer
The work done by the normal force is $0$ \, \text{J}.
Answer for screen readers
The work done by the normal force is $0$ , \text{J}.
Steps to Solve
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Understand the Work Done by Normal Force Work done is defined as the product of force and displacement in the direction of the force. The formula is given by: $$ W = F \cdot d \cdot \cos(\theta) $$ where ( W ) is work, ( F ) is the force, ( d ) is the distance, and ( \theta ) is the angle between the force and the direction of displacement.
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Identify the Normal Force For an object on a level surface with no vertical movement, the normal force ( F_n ) equals the weight of the crate: $$ F_n = m \cdot g $$ where ( m = 28.9, \text{kg} ) is the mass of the crate and ( g = 9.81, \text{m/s}^2 ) is the acceleration due to gravity.
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Calculate the Normal Force Substituting the mass into the formula for normal force: $$ F_n = 28.9 \cdot 9.81 $$
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Determine the Angle In this case, the angle ( \theta ) between the normal force and the direction of displacement is ( 90^\circ ) since the normal force acts vertically upward while the displacement is horizontal.
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Calculate Work Done Since ( \cos(90^\circ) = 0 ), the work done by the normal force is: $$ W = F_n \cdot d \cdot \cos(90^\circ) = F_n \cdot d \cdot 0 = 0 $$
The work done by the normal force is $0$ , \text{J}.
More Information
In physics, when the force and the direction of displacement are perpendicular, no work is done by that force. This is why the normal force does not contribute to the work done on the crate while it is pushed horizontally.
Tips
- Confusing the normal force with the applied force and calculating work from the applied force instead.
- Forgetting that the angle matters; if it's not along the direction of motion, it will affect the work calculation.
- Misapplying the work formula by not recognizing that the normal force contributes no work in this particular scenario.
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