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You push your physics book 1.50 m along a horizontal tabletop with a horizontal push of 2.40 N while the opposing force of friction is 0.600 N. How much work does each of the following forces do on the book: (a) your 2.40-N push, (b) the friction force, (c) the normal force from the tabletop, and (d) gravity? (e) What is the net work done on the book?

Short Answer

Expert verified
3.60 J by push, -0.900 J by friction, 0 J by normal force, 0 J by gravity, and 2.70 J net work.

Step by step solution

01

Calculate Work Done by Your Push

The work done by a force is calculated using the formula: \( W = F \times d \times \cos(\theta) \), where \( F \) is the force, \( d \) is the distance, and \( \theta \) is the angle between the force and the direction of motion. For your 2.40-N push, the force is in the same direction as the motion, so \( \theta = 0^\circ \) and \( \cos(0^\circ) = 1 \). Thus, \( W_{push} = 2.40 \times 1.50 \times 1 = 3.60 \, \text{J} \).
02

Calculate Work Done by Friction

The force of friction opposes the direction of motion, so \( \theta = 180^\circ \) and \( \cos(180^\circ) = -1 \). The work done by friction is \( W_{friction} = 0.600 \times 1.50 \times (-1) = -0.900 \, \text{J} \).
03

Calculate Work Done by the Normal Force

The normal force acts perpendicular to the direction of motion. Since \( \theta = 90^\circ \) for the normal force and \( \cos(90^\circ) = 0 \), the work done by the normal force is \( W_{normal} = 0 \).
04

Calculate Work Done by Gravity

Gravity also acts perpendicular to the horizontal direction of motion. Since \( \theta = 90^\circ \) for gravity and \( \cos(90^\circ) = 0 \), the work done by the force of gravity is \( W_{gravity} = 0 \).
05

Calculate the Net Work Done on the Book

The net work done on the book is the sum of the work done by all forces. Thus, \( W_{net} = W_{push} + W_{friction} + W_{normal} + W_{gravity} = 3.60 + (-0.900) + 0 + 0 = 2.70 \, \text{J} \).

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Key Concepts

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

Understanding Force
Force is a push or pull exerted on an object. It can change the object's speed, direction, or shape. In our exercise, there are several forces at play. These include the push you exert on the book and the opposing frictional force.
Forces are vector quantities, meaning they have both magnitude and direction. When calculating work done by these forces, it's important to know the angle between the force direction and the motion. This angle helps determine the component of the force that contributes to moving the object, as seen in the equation for work:
  • The formula for work is: \( W = F \times d \times \cos(\theta) \)
In this case, your push is fully aligned with the motion (\( \theta = 0^\circ \)), making all its magnitude effective for doing work.
The Role of Friction
Friction is a force that opposes the motion of objects sliding against each other. In our scenario, it acts to resist the movement of the book as you push it. Friction is crucial because it prevents objects from sliding indefinitely when there's no continuous external push.
When calculating work done by friction, we note that it acts in the opposite direction to the push. This means the angle \( \theta \) in the work formula becomes \( 180^\circ \). Hence, the work done by friction is negative (\( W_{friction} = -0.900 \, \text{J} \)), indicating that it takes energy away from the system.
  • Frictional work can be a vital consideration in energy budgets, where losses are significant.
Normal Force Explained
The normal force acts perpendicular to the surface in contact with an object. It is a reactive force that balances the weight of an object resting on a surface, preventing it from falling through.
In the exercise, the normal force comes from the tabletop and acts upward, perpendicular to the horizontal push. Since it doesn't have a component in the direction of motion (\( \theta = 90^\circ \)), it does no work on the book:
  • The work done is zero since \( \cos(90^\circ) = 0 \).
The normal force is crucial for determining how other forces act on objects, often appearing in calculations involving friction.
Gravity's Role
Gravity is the force that attracts objects toward the Earth. It acts vertically downward, and although prominent generally, it doesn't contribute to work in horizontal movements like our book's slide.
In this case, gravity acts at \( \theta = 90^\circ \) to the direction of motion, so it doesn't impact the work done on the book. Like the normal force, it does zero work here:
  • Gravity's fundamental impact is in vertical motion, providing energy for movement such as falling.
Understanding gravity helps in analyzing energy transitions and movements in physics.
Calculating Net Work
Net work is the total work done considering all forces acting on an object. It reveals how much energy is transferred due to these forces, accounting for opposing and supporting influences.
To calculate net work, sum up the work from all individual forces. Here, it's the combination of your push, friction, the normal force, and gravity:
  • \( W_{net} = 3.60 \text{ J} + (-0.900 \text{ J}) + 0 \text{ J} + 0 \text{ J} = 2.70 \text{ J} \)
  • Net work helps determine the resultant energy change in the object's motion.
Net work indicates whether an object accelerates, tends to move, or remains stationary based on total energy dynamics.

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Most popular questions from this chapter

An elevator has mass 600 kg, not including passengers. The elevator is designed to ascend, at constant speed, a vertical distance of 20.0 m (five floors) in 16.0 s, and it is driven by a motor that can provide up to 40 hp to the elevator. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has mass 65.0 kg.

The Grand Coulee Dam is 1270 m long and 170 m high. The electrical power output from generators at its base is approximately 2000 MW. How many cubic meters of water must flow from the top of the dam per second to produce this amount of power if 92% of the work done on the water by gravity is converted to electrical energy? (Each cubic meter of water has a mass of 1000 kg.)

A small block with a mass of 0.0600 kg is attached to a cord passing through a hole in a frictionless, horizontal surface (\(\textbf{Fig. P6.71}\)). The block is originally revolving at a distance of 0.40 m from the hole with a speed of 0.70 m/s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.10 m. At this new distance, the speed of the block is 2.80 m/s. (a) What is the tension in the cord in the original situation, when the block has speed \(\upsilon = 0.70\) m/s? (b) What is the tension in the cord in the final situation, when the block has speed \(\upsilon = 2.80\) m/s? (c) How much work was done by the person who pulled on the cord?

A proton with mass 1.67 \(\times\) 10\(^{-27}\) kg is propelled at an initial speed of 3.00 \(\times\) 10\(^5\) m/s directly toward a uranium nucleus 5.00 m away. The proton is repelled by the uranium nucleus with a force of magnitude \(F = \alpha/x^2\), where \(x\) is the separation between the two objects and \(\alpha = 2.12 \times 10^{-26} \, \mathrm{N} \cdot \mathrm{m}^2\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10}\) m from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again 5.00 m away from the uranium nucleus?

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

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