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A collie drags its bed box across a floor by applying a horizontal force of \(8.0 \mathrm{~N}\). The kinetic frictional force acting on the box has magnitude \(5.0 \mathrm{~N}\). As the box is dragged through \(0.70 \mathrm{~m}\) along the way, what are (a) the work done by the collie's applied force and (b) the increase in thermal energy of the bed and floor?

Short Answer

Expert verified
(a) 5.6 J (b) 3.5 J.

Step by step solution

01

Calculating Work Done by the Collie's Force

The work done by a force can be calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the magnitude of the force, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement. Since the force is horizontal and the displacement is also horizontal (\( \theta = 0 \)), the angle \( \theta = 0^{\circ} \) and \( \cos(0) = 1 \). Thus, the work done by the collie's force is:\[W = 8.0 \text{ N} \times 0.70 \text{ m} \times 1 = 5.6 \text{ J}\]
02

Calculating the Increase in Thermal Energy

When an object moves due to an applied force with friction present, the work done against the frictional force translates into an increase in thermal energy. The thermal energy increase is equal to the work done by the frictional force. The kinetic frictional force is \( 5.0 \text{ N} \), and the box is displaced by \( 0.70 \text{ m} \). Hence, the thermal energy increase is:\[\Delta E_{\text{thermal}} = f_{\text{friction}} \times d = 5.0 \text{ N} \times 0.70 \text{ m} = 3.5 \text{ J}\]

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Key Concepts

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

Kinetic Friction
When objects slide against each other, there is a resistance called friction. Kinetic friction is the specific type of friction that acts on moving objects. It opposes the direction of motion and is always less than or equal to the force applied. In our exercise, the kinetic frictional force between the collie's bed box and the floor amounts to \(5.0 \text{ N}\). This force attempts to stop the box from moving as the collie drags it along.

Friction depends on both the nature of the surfaces in contact and how hard they press against each other. This interaction creates a force which we calculate using the frictional force formula:
\[f_{\text{friction}} = \mu_k \times N\]
where \(\mu_k\) is the coefficient of kinetic friction and \(N\) is the normal force, or the force perpendicular to the surfaces.
  • Kinetic friction acts horizontally in this case.
  • The collie's applied force must be larger than the frictional force for movement.
  • As the object continues to move, kinetic friction transforms some mechanical energy into thermal energy.
Thermal Energy
Thermal energy is the energy that comes from heat. When we talk about friction converting into thermal energy, we're touching upon the transformation of energy from mechanical to heat. This happens because as surfaces slide over each other, microscopic irregularities cause energy to be lost in the form of heat. In our scenario, as the collie drags the box, the kinetic friction of \(5.0 \text{ N}\) results in an increase in thermal energy.


The increase in this energy can be calculated using the work done by the frictional force. Here the transformation of work into thermal energy is quantified:
\[\Delta E_{\text{thermal}} = 5.0 \text{ N} \times 0.70 \text{ m} = 3.5 \text{ J}\]

  • Thermal energy increase is equal to the work done by friction.
  • Friction continuously converts mechanical energy into heat.
  • This is why surfaces often get warm when rubbed together.
Work Done
In the world of physics, work is done when a force causes displacement. The equation \(W = F \cdot d \cdot \cos(\theta)\) helps us determine how much work is performed. In this example, the collie's force of \(8.0 \text{ N}\) is at an angle of \(0^{\circ}\) with the direction of movement, meaning all the force effectively contributes to the displacement of the box.

The work done by the collie's force is:
\[W = 8.0 \text{ N} \times 0.70 \text{ m} \times 1 = 5.6 \text{ J}\]

Things to remember about work:
  • Work is only done when there is displacement in the direction of force.
  • An angle \(\theta = 0^{\circ}\) simplifies calculations because \(\cos(0) = 1\).
  • The concept of work helps us measure the energy transferred by forces.

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Most popular questions from this chapter

A block of mass \(m=2.5 \mathrm{~kg}\) slides head on into a spring of spring constant \(k=320 \mathrm{~N} / \mathrm{m}\). When the block stops, it has compressed the spring by \(7.5 \mathrm{~cm}\). The coefficient of kinetic friction between block and floor is \(0.25\). While the block is in contact with the spring and being brought to rest, what are (a) the work done by the spring force and (b) the increase in thermal energy of the block-floor system? (c) What is the block's speed just as it reaches the spring?

The luxury liner Queen Elizabeth 2 has a diesel-electric power plant with a maximum power of \(92 \mathrm{MW}\) at a cruising speed of \(32.5\) knots. What forward force is exerted on the ship at this speed? \((1 \mathrm{knot}=1.852 \mathrm{~km} / \mathrm{h} .)\)

A \(0.63 \mathrm{~kg}\) ball thrown directly upward with an initial speed of \(14 \mathrm{~m} / \mathrm{s}\) reaches a maximum height of \(8.1 \mathrm{~m}\). What is the change in the mechanical energy of the ball-Earth system during the ascent of the ball to that maximum height?

A \(5.0 \mathrm{~kg}\) block is projected at \(5.0 \mathrm{~m} / \mathrm{s}\) up a plane that is inclined at \(30^{\circ}\) with the horizontal. How far up along the plane does the block go (a) if the plane is frictionless and (b) if the coefficient of kinetic friction between the block and the plane is \(0.40 ?\) (c) In the latter case, what is the increase in thermal energy of block and plane during the block's ascent? (d) If the block then slides back down against the frictional force, what is the block's speed when it reaches the original projection point?

A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?

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