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Chapter 7: Problem 89

A boy pulls a 5-kg cart with a 20-N force at an angle of \(30^{\circ}\) above the horizontal for a length of time. Over this time frame, the cart moves a distance of \(12 \mathrm{m}\) on the horizontal floor. (a) Find the work done on the cart by the boy. (b) What will be the work done by the boy if he pulled with the same force horizontally instead of at an angle of \(30^{\circ}\) above the horizontal over the same distance?

Short Answer

Expert verified
(a) The work done by the boy while pulling the cart at an angle of \(30^{\circ}\) is approximately \(208.32 \mathrm{J}\). (b) The work done by the boy while pulling the cart horizontally is \(240 \mathrm{J}\).

Step by step solution

01

Calculate the horizontal component of force

Since the force is acting at an angle, we can find the horizontal component of the force, Fx, by using Fx = F * cos胃, where F is the force (20 N), and 胃 is the angle (30掳). In this case: Fx = 20 * cos(30掳)
02

Calculate the work done on the cart

Now that we have the horizontal component of the force, we can calculate the work done by the boy by using the formula W = Fx * d, where W is the work, Fx is the horizontal force component, and d is the distance (12 m). In this case: W = Fx * 12 Plug in the value of Fx we calculated before: W = (20 * cos(30掳)) * 12
03

Solve for the work done on the cart

After plugging in all the values, simplify and solve for W: \( W = 20 * 0.866 * 12 \) \( W = 208.32 \) The work done by the boy while pulling the cart at an angle of 30掳 is approximately 208.32 J (Joules).
04

Calculate the work done if force applied horizontally

In this case, when the boy pulls horizontally, the angle 胃 is 0掳. Now we can use the same formula but with a different angle. As the force applied horizontally, we don't need to calculate the horizontal component of the force. So, we can directly move to the work formula: W = F * d * cos(0掳), where W is the work, F is the force (20 N), and d is the distance (12 m): W = 20 * 12 * cos(0掳)
05

Solve for the work done if force applied horizontally

After plugging in all the values, simplify and solve for W: \( W = 20 * 12 * 1 \) \( W = 240 \) The work done by the boy while pulling the cart horizontally is 240 J (Joules). To sum up, (a) the work done by the boy while pulling the cart at an angle of 30掳 is approximately 208.32 J and (b) the work done by the boy while pulling the cart horizontally is 240 J.

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Key Concepts

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

Force
Force is a fundamental concept in physics, describing an interaction that changes an object's motion or shape. The force exerted by the boy pulling the cart is crucial for calculating the work done. In this exercise, the boy applies a force of 20 Newtons to the cart. A Newton, abbreviated as N, is the standard unit of force. Recall that force is a vector quantity, which means it has both a magnitude (size) and direction. These two attributes are essential for understanding how force affects the movement of an object, especially when the force is applied at an angle.
Work Done
Work done refers to the amount of energy transferred by a force acting on an object as it moves through a distance. It can be calculated using the formula \( W = F \, \times \, d \, \times \, \cos(\theta) \), where \( W \) is the work done, \( F \) is the force applied, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion.
  • When the force is aligned with the object's motion, work is maximized (as seen when the boy pulls horizontally).
  • If the force acts at an angle, the work done decreases, because only a part of the force contributes to moving the object in its direction.
This exercise shows that pulling at an angle reduces the effectiveness of the force in doing work, as noted in the different values of 208.32 J and 240 J for work done at different angles.
Angle of Force
The angle of force is the angle at which the force is applied relative to a reference line, often horizontal. In this problem, the boy pulls the cart with a force that makes an angle of \(30^{\circ}\) with the horizontal.
The angle can significantly impact the horizontal component of the force, altering how much of the actual force contributes to moving the object straight ahead. The effect of different angles is evident in the solutions, where pulling horizontally (\(0^{\circ}\) angle) allows the entire force to be applied along the motion, yielding more work.
Horizontal Component
The horizontal component of the force is the part of the force acting parallel to the horizontal surface. This component is crucial when the force is applied at an angle, as only this portion contributes to moving the object forward.
To calculate the horizontal component, we use the formula \( F_{x} = F \, \times \, \cos(\theta) \). In this scenario:
  • The total force \( F \) is 20 N.
  • The angle \( \theta \) is \(30^{\circ} \).
  • Thus, \( F_{x} = 20 \, \times \, \cos(30^{\circ}) \), which simplifies to approximately 17.32 N.
This value is then used to determine the work done by calculating \( W = F_{x} \, \times \, 12 \). Understanding the horizontal component ensures you can correctly calculate how each part of the force contributes to doing work.

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Most popular questions from this chapter

(a) How fast must a 3000-kg elephant move to have the same kinetic energy as a 65.0 -kg sprinter running at \(10.0 \mathrm{m} / \mathrm{s}\) ? (b) Discuss how the larger energies needed for the movement of larger animals would relate to metabolic rates.

Calculate the work done by an 85.0 -kg man who pushes a crate \(4.00 \mathrm{m}\) up along a ramp that makes an angle of \(20.0^{\circ}\) with the horizontal (see below). He exerts a force of \(500 \mathrm{N}\) on the crate parallel to the ramp and moves at a constant speed. Be certain to include the work he does on the crate and on his body to get up the ramp.

Shown below is a 40-kg crate that is pushed at constant velocity a distance \(8.0 \mathrm{m}\) along a \(30^{\circ}\) incline by the horizontal force \(\overrightarrow{\mathbf{F}}\). The coefficient of kinetic friction between the crate and the incline is \(\mu_{k}=0.40 .\) Calculate the work done by (a) the applied force, (b) the frictional force, (c) the gravitational force, and (d) the net force.

A girl pulls her 15-kg wagon along a flat sidewalk by applying a \(10-\mathrm{N}\) force at \(37^{\circ}\) to the horizontal. Assume that friction is negligible and that the wagon starts from rest. (a) How much work does the girl do on the wagon in the first \(2.0 \mathrm{s}\). (b) How much instantaneous power does she exert at \(t=2.0 \mathrm{s} ?\)

Suppose a car travels \(108 \mathrm{km}\) at a speed of \(30.0 \mathrm{m} / \mathrm{s}\), and uses 2.0 gal of gasoline. Only \(30 \%\) of the gasoline goes into useful work by the force that keeps the car moving at constant speed despite friction. (The energy content of gasoline is about 140 MJ/gal.) (a) What is the magnitude of the force exerted to keep the car moving at constant speed? (b) If the required force is directly proportional to speed, how many gallons will be used to drive \(108 \mathrm{km}\) at a speed of \(28.0 \mathrm{m} / \mathrm{s} ?\)
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