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Chapter 7: Problem 54

A 5.0-kg box has an acceleration of \(2.0 \mathrm{m} / \mathrm{s}^{2}\) when it is pulled by a horizontal force across a surface with \(\mu_{K}=0.50 .\) Find the work done over a distance of \(10 \mathrm{cm}\) by (a) the horizontal force, (b) the frictional force, and (c) the net force. (d) What is the change in kinetic energy of the box?

Short Answer

Expert verified
The work done by the horizontal force is 3.4525 J, the work done by the frictional force is -2.4525 J, the work done by the net force is 1.0 J, and the change in kinetic energy of the box is 1.0 J.

Step by step solution

01

Calculate the normal force on the box

Since there is no vertical acceleration, the normal force (N) is equal to the gravitational force acting on the box (Fg = mg). So, N = mg \(N = (5.0 kg)(9.81 \frac{m}{s^2})\) N = 49.05 N
02

Calculate the frictional force

We can find the frictional force (F_friction) using the equation: F_friction = 碌k * N F_friction = 0.50 * 49.05 N F_friction = 24.525 N
03

Calculate the horizontal force

We can find the horizontal force (F_horizontal) required to accelerate the box using the equation: F_horizontal = m * a + F_friction F_horizontal = (5.0 kg * 2.0 \(\frac{m}{s^2}\)) + 24.525 N F_horizontal = 34.525 N
04

Calculate the work done by the horizontal force, the frictional force, and the net force

Work is defined as the force times the distance over which the force is applied. First, let's convert the distance from cm to m: distance = 10 cm = 0.1 m Now, we can calculate the work done by each force: W_horizontal = F_horizontal * distance W_horizontal = 34.525 N * 0.1 m W_horizontal = 3.4525 J W_friction = -F_friction * distance (the negative sign indicates that the force acts in the opposite direction) W_friction = -24.525 N * 0.1 m W_friction = -2.4525 J W_net = (F_horizontal + F_friction) * distance W_net = (34.525 N - 24.525 N) * 0.1 m W_net = 1.0 J
05

Calculate the change in kinetic energy

The work-energy theorem states that the work done by the net force is equal to the change in kinetic energy. Therefore, 螖KE = W_net 螖KE = 1.0 J Thus, the work done by the horizontal force is 3.4525 J, the work done by the frictional force is -2.4525 J, the work done by the net force is 1.0 J, and the change in kinetic energy of the box is 1.0 J.

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Key Concepts

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

Normal Force
In many physics problems, understanding the concept of normal force is crucial. The normal force is a supportive force exerted by a surface that keeps objects resting on it from falling through. It acts perpendicular to the surface. Since the box in our problem is on a horizontal surface and not accelerating vertically, its normal force equals the gravitational force.
  • The force of gravity acting downwards is given by the equation: \[ F_{gravity} = mg \] where \( m \) is the mass of the object (5.0 kg in this case), and \( g \) is the acceleration due to gravity (approx. \( 9.81 \, \mathrm{m/s^2} \)). This results in a normal force of \( 49.05 \, \mathrm{N} \).
  • This force ensures that the box remains stationary unless another force acts on it horizontally.
Understanding the normal force is essential, as it directly impacts the calculation of frictional forces, which we'll explore next.
Frictional Force
Frictional force is the resistance that one surface or object encounters when moving over another. It acts opposite to the direction of motion. In this scenario, the box is moving over a surface with a kinetic friction coefficient (\( \mu_k = 0.50 \)). The frictional force is calculated using the normal force:
  • Frictional force (\( F_{friction} \)) can be expressed by the equation:\[ F_{friction} = \mu_k \cdot N \]
  • Substituting the given values: \[ F_{friction} = 0.50 \times 49.05 \, \mathrm{N} = 24.525 \, \mathrm{N} \]
  • This force resists the box's movement, requiring the horizontal force to overcome it for acceleration to occur.
Friction plays a critical role in determining the total work done and affects how forces are balanced as objects move.
Work-Energy Theorem
The work-energy theorem is a foundational concept that links the work done on an object to changes in its kinetic energy. It states that the total work done by all forces acting on an object equals the change in its kinetic energy.
  • This theorem can be mathematically represented as:\[ W_{net} = \Delta KE \]
  • In our box scenario:
    • The net work done (\( W_{net} \)) is calculated as \( 1.0 \, \mathrm{J} \), which corresponds to the change in kinetic energy during its movement over the \( 10 \, \mathrm{cm} \) distance.
    • This means the energy from the work done by the net force has converted into an increase in the box's kinetic energy.
Understanding how work translates into energy changes helps us predict system behavior under applied forces.
Kinetic Energy Change
Kinetic energy, the energy of motion, depends greatly on the mass and velocity of an object. The change in kinetic energy (\( \Delta KE \)) during an object's motion tells us how much its speed has altered due to work done on it.
  • The formula for kinetic energy (KE) is:\[ KE = \frac{1}{2}mv^2 \]
  • In this exercise:
    • Initially, the box had some velocity, and after the force was applied and work was done, the kinetic energy increased.
    • The calculated change in kinetic energy is \( 1.0 \, \mathrm{J} \), directly derived from the net work done over the surface.
This change reflects the box's accelerated motion, influenced by the forces applied, comprehending how forces convert to movement and energy efficiency.

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