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Chapter 11: Problem 3

To push a \(25-\mathrm{kg}\) crate up a \(27^{\circ}\) incline, a worker exerts a force of \(120 \mathrm{~N}\), parallel to the incline. As the crate slides \(3.6 \mathrm{~m}\), how much work is done on the crate by \((a)\) the worker, \((b)\) the force of gravity, and \((c)\) the normal force of the incline?

Short Answer

Expert verified
The work done by the worker is \(432 \mathrm{~J}\), by the force of gravity is \(-675 \mathrm{~J}\) and by the normal force of the incline is \(0 \mathrm{~J}\).

Step by step solution

01

Calculation of work done by the worker

The work done by the worker is calculated using the formula: \(Work = Force \times Distance\). Here, the force is given as \(120 \mathrm{~N}\) and the distance is \(3.6 \mathrm{~m}\), thus, \(Work_{worker} = 120 \mathrm{~N} \times 3.6 \mathrm{~m} = 432 \mathrm{~J}\).
02

Calculation of work done by the force of gravity

The work done by gravity is calculated using the formula: \(Work = Force \times Distance \times cos(\Theta)\). Here, the force is the weight of the crate, which is \(25 \mathrm{kg} \times 9.8 m/s^2 = 245 \mathrm{~N}\). The angle of incline is \(27^\circ\), therefore, \(Work_{gravity} = 245 \mathrm{~N} \times 3.6 \mathrm{~m} \times cos(27^\circ) = -675 \mathrm{~J}\). The negative sign indicates that the work done by gravity is in the opposite direction to the motion of the crate.
03

Calculation of work done by the normal force

The normal force acts perpendicular to the direction of motion, and the work done by a force perpendicular to the direction of motion is always zero. Thus, \(Work_{normal} = 0\).
04

Summarize the results and make conclusions

The work done by the worker on the crate is \(432 \mathrm{~J}\), by the gravity is \(-675 \mathrm{~J}\) and by the normal force is \(0 \mathrm{~J}\). The negative sign in the work done by gravity indicates that this work opposes the motion of the crate.

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Most popular questions from this chapter

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