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Difficulty: MediumWork, Energy, and Power

A technician performs 750 J750\text{ J} of mechanical work while pushing a piece of machinery across a level workshop floor over a distance of 15 meters15\text{ meters}. Assuming the force is applied parallel to the direction of motion, what is the magnitude of the horizontal force applied, in newtons?

Answer: 50 N

Answer

The magnitude of the applied horizontal force is 50 N50\text{ N}.
Mechanical work is defined as W=F×dW = F \times d when force and displacement are in the same direction. Rearranging to solve for force gives F=Wd=750 J15 m=50 NF = \frac{W}{d} = \frac{750\text{ J}}{15\text{ m}} = 50\text{ N}.

Step-by-Step Solution

1
State the governing formula relating work, force, and displacement
W=F×dW = F \times d
Work is defined as force multiplied by displacement in the direction of the force.
2
Rearrange the equation to isolate the unknown force variable (FF)
F=WdF = \frac{W}{d}
Dividing both sides of the work formula by displacement yields force.
3
Substitute the given numerical quantities (W=750 JW = 750\text{ J}, d=15 md = 15\text{ m}) and compute
F=75015=50 NF = \frac{750}{15} = 50\text{ N}
Carrying out the arithmetic yields the magnitude of the required force in newtons.

Key Concept

Work is the product of the magnitude of displacement and the component of force in the direction of displacement (W=FdW = Fd).
Estimated Time:1m 0s
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Topics

Work, Energy, and Power

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