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Chapter 5: Problem 52

How much work is done against gravity in lifting a \(6.00-\mathrm{kg}\) weight through a distance of \(20.0 \mathrm{~cm} ?\)

Short Answer

Expert verified
Answer: The work done against gravity is 11.772 J.

Step by step solution

01

Convert distance

First, we need to convert the given distance from centimeters to meters, since we'll be working in SI units. To do this, remember that there are \(100\) centimeters in a meter: \(20.0\mathrm{~cm} = 20.0/100\mathrm{~m} = 0.200\mathrm{~m}\).
02

Calculate gravitational force

Next, we need to find the gravitational force acting on the weight. This is given by the formula \(F = m \times g\), where \(m\) is the mass of the weight and \(g\) is the gravitational acceleration (\(9.81\mathrm{~m/s^2}\)). So the gravitational force is: \(F = 6.00\mathrm{~kg} \times 9.81\mathrm{~m/s^2} = 58.86\mathrm{~N}\).
03

Compute work done against gravity

Now that we have the distance and the gravitational force, we can calculate the work done against gravity using the formula for work: \(W = F \times d \times \cos{\theta}\). Since the angle is \(0^{\circ}\), the cosine of the angle is \(1\). Therefore, the work done against gravity is: \(W = 58.86\mathrm{~N} \times 0.200\mathrm{~m} \times 1 = 11.772 \mathrm{~J}\) (joules). So, the work done against gravity in lifting the \(6.00-\mathrm{kg}\) weight through a distance of \(20.0\mathrm{~cm}\) is \(11.772\mathrm{~J}\).

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Most popular questions from this chapter

An arrow of mass \(m=88 \mathrm{~g}(0.088 \mathrm{~kg})\) is fired from a bow. The bowstring exerts an average force of \(F=110 \mathrm{~N}\) on the arrow over a distance \(d=78 \mathrm{~cm}(0.78 \mathrm{~m})\) Calculate the speed of the arrow as it leaves the bow.

An engine expends 40.0 hp in moving a car along a level track at a speed of \(15.0 \mathrm{~m} / \mathrm{s}\). How large is the total force acting on the car in the opposite direction of the motion of the car?

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) \(\mathrm{m} / \mathrm{s}^{2}\)

An advertisement claims that a certain \(1200-\mathrm{kg}\) car can accelerate from rest to a speed of \(25 \mathrm{~m} / \mathrm{s}\) in \(8.0 \mathrm{~s}\). What average power must the motor supply in order to cause this acceleration? Ignore losses due to friction.

A car of mass 942.4 kg accelerates from rest with a constant power output of 140.5 hp. Neglecting air resistance, what is the speed of the car after 4.55 s?
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