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Chapter 6: Problem 1

An old oaken bucket of mass 6.75 \(\mathrm{kg}\) hangs in a well at the end of a rope. The rope passes over a frictionless pulley at the top of the well, and you pull horizontally on the end of the rope to raise the bucket slowly a distance of 4.00 \(\mathrm{m} .\) (a) How much work do you do on the bucket in pulling it up? (b) How much work does gravity do on the bucket? (c) What is the total work done on the bucket?

Short Answer

Expert verified
(a) 0 J, (b) -264.6 J, (c) -264.6 J.

Step by step solution

01

Identify the Forces Involved

To solve this problem, we first need to identify the forces acting on the bucket as it is being pulled up. The main forces to consider are the force of gravity acting downwards and the force exerted by you pulling horizontally on the rope to lift the bucket.
02

Calculate the Work Done by You

Work is calculated using the formula \( W = F \, d \, \cos(\theta) \), where \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of motion. In this case, since you are pulling horizontally and the bucket moves vertically, \( \theta = 90^\circ \) and \( \cos(90^\circ) = 0 \). Therefore, the work done by you is \( 0 \ \mathrm{J} \) because the force you apply horizontally doesn't move the bucket upwards.
03

Calculate the Work Done by Gravity

The work done by gravity is calculated using the same work formula: \( W = F \, d \, \cos(\theta) \). The force due to gravity is \( F_g = m \, g \), where \( m = 6.75 \, \mathrm{kg} \) and \( g = 9.8 \, \mathrm{m/s^2} \). Thus, \( F_g = 6.75 \, \mathrm{kg} \times 9.8 \, \mathrm{m/s^2} = 66.15 \, \mathrm{N} \). The work done by gravity, with \( \theta = 180^\circ \), is \( W = 66.15 \, \mathrm{N} \times 4.0 \, \mathrm{m} \times (-1) = -264.6 \, \mathrm{J} \). The negative sign indicates that gravity does negative work as it opposes the upward movement.
04

Determine the Total Work Done on the Bucket

The total work done on the bucket is the sum of the work done by you and the work done by gravity. Therefore, the total work done is \( 0 \, \mathrm{J} + (-264.6 \, \mathrm{J}) = -264.6 \, \mathrm{J} \). This means that overall, there is negative work done on the bucket, as gravity is doing most of the work against the upward movement.

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Key Concepts

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

Forces in Physics
Forces are fundamental to understanding how objects interact in our universe. In this exercise, two main forces act on the oaken bucket: gravity and the force you apply to the rope. Let's break these down:

Gravity is a force that pulls objects towards the center of the Earth. It's always acting downwards and has a magnitude calculated by multiplying the object's mass by the acceleration due to gravity (9.8 m/s虏).

The force you apply on the rope is horizontal. However, since the bucket needs to move vertically, this horizontal force doesn't contribute to lifting the bucket. It's crucial to understand that forces can act in different directions, and not all applied forces contribute to motion. This is why the angle between the force applied and the direction of movement is essential in calculating work.

  • **Gravity Force**: Always acts downwards with magnitude given by mass times gravitational acceleration.
  • **Applied Force**: In this case acts horizontally and does not contribute to vertical motion because it is perpendicular to it.
Work Done by Gravity
Gravitational work involves the interaction between the gravitational force and the movement of an object. When lifting the bucket, gravity pulls it downward, doing negative work. Negative work means that the force is acting in the opposite direction to the displacement.

The formula used to calculate work from gravity is: \( W = F \cdot d \cdot \cos(\theta) \)
where:
  • \( F \) is the force due to gravity.
  • \( d \) is the displacement (4 m, vertically upward).
  • \( \theta \) for gravity here is 180掳, because the force of gravity opposes the direction of the bucket's movement.

This calculation shows that 264.6 J of work is done negatively by gravity as it tries to pull the bucket back down. Understanding this helps illustrate why more effort must be exerted to move against gravity.
Physics Problem Solving
Physics problems, like lifting the bucket, require understanding physical laws and correctly applying mathematical formulas. Here's a step-by-step approach to tackling such problems:

1. **Identify the Forces**: Know what forces are at play, their directions, and magnitudes. In this exercise, it's gravity and the horizontal pull on the rope.

2. **Use the Proper Formula**: Recall the work formula \( W = F \cdot d \cdot \cos(\theta) \). This helps calculate the work done by each force. Remember, work is only done if the force moves the object in the direction of force or has a component in the direction of motion.

3. **Consider Angles Carefully**: The angle \( \theta \) plays a key role. If the force and movement direction are perpendicular (90掳), the work done is zero. If opposed, like gravity here, it's negative.

4. **Calculate Accurately**: For applied problems, substituting the correct values into the equations ensures accurate results. Always check units and recalculate when necessary.

By breaking down problems into these steps, students can systematically solve complex problems while gaining a deeper understanding of physics principles.

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Most popular questions from this chapter

On a winter's day in Maine, a warehouse worker is shoving boxes up a rough plank inclined at an angle \(\alpha\) above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance \(x\) along the plank: \(\mu=A x\), where \(A\) is a positive constant and the bottom of the plank is at \(x=0 .\) (For this plank the coefficients of kinetic and static friction are equal: \(\mu_{k}=\mu_{s}=\mu\) . The worker shoves a box up the plank so that it leaves the bottom of the plank moving at speed \(v_{0}\) . Show that when the box first comes to rest, it will remain at rest if $$v_{0}^{2} \geq \frac{3 g \sin ^{2} \alpha}{A \cos \alpha}$$

You and your bicycle have combined mass 80.0 \(\mathrm{kg}\) . When you reach the base of a bridge, you are traveling along the road at 5.00 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} .6 .35) .\) At the top of the bridge, you have climbed a vertical distance of 5.20 \(\mathrm{m}\) and have slowed to 1.50 \(\mathrm{m} / \mathrm{s}\) . You can ignore work done by friction and any inefficiency in the bike or your legs. (a) What is the total work done on you and your bicycle when you go from the base to the top of the bridge? (b) How much work have you done with the force you apply to the pedals?

A soccer ball with mass 0.420 \(\mathrm{kg}\) is initially moving with speed 2.00 \(\mathrm{m} / \mathrm{s}\) . A soccer player kicks the ball, exerting a constant force of magnitude 40.0 \(\mathrm{N}\) in the same direction as the ball's motion. Over what distance must the player's foot be in contact with the ball to increase the ball's speed to 6.00 \(\mathrm{m} / \mathrm{s} ?\)

A luggage handler pulls a \(20.0-\mathrm{kg}\) suitcase up a ramp inclined at \(25.0^{\circ}\) above the horizontal by a force \(\overrightarrow{\boldsymbol{F}}\) of magnitude 140 \(\mathrm{N}\) that acts parallel to the ramp. The coefficient of kinetic friction between the ramp and the incline is \(\mu_{\mathrm{k}}=0.300\) . If the suitcase travels 3.80 \(\mathrm{m}\) along the ramp, calculate (a) the work done on the suitcase by the force \(\overrightarrow{\boldsymbol{F}},\) (b) the work done on the suitcase by the gravitational force; (c) the work done on the suitcase by the normal force; (d) the work done on the suitcase by the friction force; (c) the total work done on the suitcase. (f) If the speed of the suitcase is zero at the bottom of the ramp, what is its speed after it has traveled 3.80 \(\mathrm{m}\) m along the ramp?

A block of ice with mass 6.00 \(\mathrm{kg}\) is initially at rest on a frictionless, horizontal surface. A worker then applies a horizontal force \(\vec{F}\) to it. As a result, the block moves along the \(x\) -axis such that its position as a function of time is given by \(x(t)=\alpha t^{2}+\beta t^{3}\) , where \(\alpha=0.200 \mathrm{m} / \mathrm{s}^{2}\) and \(\beta=0.0200 \mathrm{m} / \mathrm{s}^{3}\) . (a) Calculate the velocity of the object when \(t=4.00 \mathrm{s}\) . (b) Calculate the magnitude of \(\overrightarrow{\boldsymbol{F}}\) when \(t=4.00 \mathrm{s}\) . (c) Calculate the work done by the force \(\overrightarrow{\boldsymbol{F}}\) during the first 4.00 \(\mathrm{s}\) of the motion.
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