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Chapter 6: Problem 5

A 75.0 -kg painter climbs a ladder that is 2.75 \(\mathrm{m}\) long leaning against a vertical wall. The ladder makes a \(30.0^{\circ}\) angle with the wall. (a) How much work does gravity do on the painter? (b) Does the answer to part (a) depend on whether the painter climbs at constant speed or accelerates up the ladder?

Short Answer

Expert verified
(a) Gravity does -1009.125 J of work on the painter. (b) No, the work by gravity does not depend on the rate of climb.

Step by step solution

01

Determine the Height Gained by the Painter

First, we need to calculate how high the painter climbs. The painter only moves vertically in the context of gravity's work. Given the ladder's length of 2.75 m and an angle of \(30.0^{\circ}\) with the wall, use the sine function to find the height gained: \[ h = 2.75 \times \sin(30.0^{\circ}) \]Recall that \(\sin(30.0^{\circ}) = 0.5\), so:\[ h = 2.75 \times 0.5 = 1.375 \text{ m} \]
02

Calculate the Gravitational Force on the Painter

The gravitational force acting on the painter is given by the equation:\[ F_g = m \times g \]where \( m = 75.0 \text{ kg} \) is the mass of the painter and \( g = 9.8 \text{ m/s}^2 \) is the acceleration due to gravity. Thus:\[ F_g = 75.0 \times 9.8 = 735 \text{ N} \]
03

Compute the Work Done by Gravity

The work done by gravity is given by the equation:\[ W = F \cdot d \cdot \cos(\theta) \]Here, \( F = 735 \text{ N} \) is the gravitational force, \( d = 1.375 \text{ m} \) is the height gained, and \( \theta = 180^{\circ} \) because gravity acts downward while the painter climbs upward. Thus, \( \cos(180^{\circ}) = -1 \).Calculating the work:\[ W = 735 \times 1.375 \times (-1) = -1009.125 \text{ J} \]The negative sign indicates that gravity does negative work as it opposes the motion of the painter climbing up.
04

Consider the Impact of Painter's Speed

The work done by gravity depends solely on the change in vertical height and gravitational force. This is independent of whether the painter climbs with constant speed or accelerates because work calculated from gravitational force and vertical displacement involves no time component.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gravitational Force
Gravitational force is a fundamental concept in physics that describes the attractive force acting between two masses. In our painter scenario, it's the force that pulls the painter towards the center of the Earth. Gravity acts vertically downward and is directly proportional to the mass of the object and the acceleration due to gravity.
The gravitational force formula is given by:
  • \( F_g = m imes g \)
where:
  • \( m \) is the mass of the object (in kilograms), and
  • \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \) on Earth).
In the example provided, the painter has a mass of 75.0 kg, making the gravitational force acting on him \( 735 \text{ N} \). Understanding gravitational force helps us calculate the work gravity does on an object when it moves in a vertical direction.
Vertical Displacement
Vertical displacement is another important aspect of understanding work done by gravity. It refers to the change in height as an object moves vertically. In the painter's scenario, the vertical displacement is calculated by finding the height he gains while climbing the ladder.
This is done using trigonometric functions applied to the ladder's length and angle with the wall. Specifically, the sine function is used when dealing with right triangles:
  • \( h = L \times \sin(\theta) \)
where:
  • \( L \) is the ladder's length, and
  • \( \theta \) is the angle between the ladder and the wall.
In the exercise, the painter climbs up to a height of 1.375 meters. Knowing the displacement allows us to determine how much work gravity does as it impacts energy change when moving against or with gravitational force.
Constant Speed vs Acceleration
When calculating work done by gravity, it's essential to understand that the concept doesn't depend on how fast or slow the painter climbs the ladder. Whether the painter moves at a constant speed or accelerates, the work done by gravity remains the same.
Here's why:
  • Work is defined as the product of force and displacement (\( W = F \cdot d \cdot \cos(\theta) \)).
  • The formula doesn't include time or speed directly; it focuses only on vertical displacement and force direction.
Therefore, even if the painter climbs faster (accelerates) or maintains a steady pace (constant speed), the gravitational work, calculated by multiplying gravitational force, vertical displacement, and direction (\( \cos(180^{\circ}) = -1 \)), remains unaffected by his motion style. Understanding this helps learners focus on figuring out displacement and force to assess gravitational work.

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Most popular questions from this chapter

A crate on a motorized cart starts from rest and moves with a constant eastward acceleration of \(a=2.80 \mathrm{m} / \mathrm{s}^{2}\) . A worker assists the cart by pushing on the crate with a force that is eastward and has magnitude that depends on time according to \(F(t)=\) \((5.40 \mathrm{N} / \mathrm{s}) t .\) What is the instantaneous power supplied by this force at \(t=5.00 \mathrm{s} ?\)

BIO Should You Walk or Run? It is 5.0 \(\mathrm{km}\) from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10 \(\mathrm{km} / \mathrm{h}\) (which uses up energy at the rate of 700 \(\mathrm{W}\) ), or you could walk it leisurely at 3.0 \(\mathrm{km} / \mathrm{h}\) (which uses energy at 290 \(\mathrm{W}\) W). Which choice would burn up more energy, and how much energy (in joules) would it burn? Why is it that the more intense exercise actually burns up less energy than the less intense exercise?

At a waterpark, sleds with riders are sent along a slippery, horizontal surface by the release of a large compressed spring. The spring with force constant \(k=40.0 \mathrm{N} / \mathrm{cm}\) and negligible mass rests on the frictionless horizontal surface. One end is in contact with a stationary wall. A sled and rider with total mass 70.0 \(\mathrm{kg}\) are pushed against the other end, compressing the spring 0.375 \(\mathrm{m}\) . The sled is then released with zero initial velocity. What is the sled's speed when the spring (a) returns to its uncompressed length and (b) is still compressed 0.200 \(\mathrm{m} ?\)

A 30.0 -kg crate is initially moving with a velocity that has magnitude 3.90 \(\mathrm{m} / \mathrm{s}\) in a direction \(37.0^{\circ}\) west of north. How much work must be done on the crate to change its velocity to 5.62 \(\mathrm{m} / \mathrm{s}\) in a direction \(63.0^{\circ}\) south of east?

Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?
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