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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

A car does the work \(W_{\text {car }}=7.0 \cdot 10^{4} \mathrm{~J}\) in traveling a distance \(x=2.8 \mathrm{~km}\) at constant speed. Calculate the average force \(F\) (from all sources) acting on the car in this process.

A horse draws a sled horizontally across a snowcovered field. The coefficient of friction between the sled and the snow is \(0.195,\) and the mass of the sled, including the load, is \(202.3 \mathrm{~kg}\). If the horse moves the sled at a constant speed of \(1.785 \mathrm{~m} / \mathrm{s}\), what is the power needed to accomplish this?

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 \(1214.5 \mathrm{~kg}\) is moving at a speed of \(62.5 \mathrm{mph}\) when it misses a curve in the road and hits a bridge piling. If the car comes to rest in \(0.236 \mathrm{~s}\), how much average power (in watts) is expended in this interval?

A ski jumper glides down a \(30.0^{\circ}\) slope for \(80.0 \mathrm{ft}\) before taking off from a negligibly short horizontal ramp. If the jumper's takeoff speed is \(45.0 \mathrm{ft} / \mathrm{s}\), what is the coefficient of kinetic friction between skis and slope? Would the value of the coefficient of friction be different if expressed in SI units? If yes, by how much would it differ?
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