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Chapter 5: Problem 22

A force of \(5.00 \mathrm{~N}\) acts through a distance of \(12.0 \mathrm{~m}\) in the direction of the force. Find the work done.

Short Answer

Expert verified
Answer: The work done by the force over the distance is 60.0 J.

Step by step solution

01

Write down the given values and formula

We are given: - Force, \(F = 5.00 \mathrm{~N}\) - Distance, \(d = 12.0 \mathrm{~m}\) - Angle, \(\theta = 0^\circ\) The formula for work done is: \(W = Fd\cos{\theta}\)
02

Calculate the cosine of the angle

As the angle is \(0^\circ\), the cosine of the angle is: \(\cos{0^\circ} = 1\)
03

Plug in the values into the formula and calculate the work done

Using the values and the cosine of the angle, the work done can be calculated as: \(W = (5.00 \mathrm{~N})(12.0 \mathrm{~m})(1)\) \(W = 60.0 \mathrm{~J}\) So, the work done by the force over the distance is \(60.0 \mathrm{~J}\).

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Most popular questions from this chapter

Two cars are moving. The first car has twice the mass of the second car but only half as much kinetic energy. When both cars increase their speed by \(5.0 \mathrm{~m} / \mathrm{s}\), they then have the same kinetic energy, Calculate the original speeds of the two cars.

Think about the answers to these questions next time you are driving a car: a) What is the kinetic energy of a \(1500 .-\mathrm{kg}\) car moving at \(15.0 \mathrm{~m} / \mathrm{s} ?\) b) If the car changed its speed to \(30.0 \mathrm{~m} / \mathrm{s}\), how would the value of its kinetic energy change?

A car, of mass \(m,\) traveling at a speed \(v_{1}\) can brake to a stop within a distance \(d\). If the car speeds up by a factor of \(2, v_{2}=2 v_{1},\) by what factor is its stopping distance increased, assuming that the braking force \(F\) is approximately independent of the car's speed?

A skydiver is subject to two forces: gravity and air resistance. Falling vertically, she reaches a constant terminal speed at some time after jumping from a plane. Since she is moving at a constant velocity from that time until her chute opens, we conclude from the work-kinetic energy theorem that, over that time interval, a) the work done by gravity is zero. b) the work done by air resistance is zero. c) the work done by gravity equals the negative of the work done by air resistance. d) the work done by gravity equals the work done by air resistance. e) her kinetic energy increases.

Which of the following is a correct unit of power? a) \(\mathrm{kg} \mathrm{m} / \mathrm{s}^{2}\) c) J e) \(W\) b) \(N\) d) \(\mathrm{m} / \mathrm{s}^{2}\)
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