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Chapter 7: Problem 1

A block of mass 2.50 kg is pushed 2.20 m along a frictionless horizontal table by a constant 16.0-N force directed 25.0掳 below the horizontal. Determine the work done on the block by (a) the applied force, (b) the normal force exerted by the table, and (c) the gravitational force. (d) Determine the total work done on the block.

Short Answer

Expert verified
The work done on the block by the applied force is approximately 31.2 J, by the normal force is 0 J, by the gravitational force is 0 J, and the total work done on the block is approximately 31.2 J.

Step by step solution

01

Calculate the work done by the applied force

Work done by a force is given by the formula: \( W = F \times d \times \text{cos}(\theta) \), where \(W\) is the work, \(F\) is the magnitude of the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and the displacement vector. In this case, \(F = 16.0 \text{N}\), \(d = 2.20 \text{m}\), and \(\theta = 25.0掳\). Therefore, the work done by the applied force is \( W_{\text{applied}} = 16.0 \times 2.20 \times \text{cos}(25.0掳) \).
02

Calculate the work done by the normal force exerted by the table

The work done by the normal force is zero because the normal force is perpendicular to the displacement of the block. Since the cosine of 90掳 is zero, no work is done by the normal force (\(W_{\text{normal}} = F_{\text{normal}} \times d \times \text{cos}(90掳) = 0\)).
03

Calculate the work done by the gravitational force

The work done by the gravitational force is also zero because the gravitational force (weight of the block) is also perpendicular to the horizontal displacement of the block, thus \(W_{\text{gravitational}} = F_{\text{gravitational}} \times d \times \text{cos}(90掳) = 0\).
04

Determine the total work done on the block

The total work done on the block is the sum of the work done by all forces. As the work done by the normal force and the gravitational force is zero, the total work done is the work done by the applied force alone. \(W_{\text{total}} = W_{\text{applied}} + W_{\text{normal}} + W_{\text{gravitational}} = W_{\text{applied}}\).

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Key Concepts

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

Work-Energy Principle
Understanding the work-energy principle is crucial when studying the dynamics of objects in physics. This principle states that the work done on an object is equal to the change in its kinetic energy. That is, when work is done on an object, it either gains or loses kinetic energy, depending on the direction of the force applied relative to the object's movement.

In our exercise, we calculated the work done by a force acting at an angle to the displacement. When the force has a component in the direction of the object's motion, as in the case of the applied force on the block, it does work that changes the block's kinetic energy. This is reflected in the formula for work, \( W = F \times d \times \cos(\theta) \) where \(F\) is the magnitude of the force, \(d\) is the displacement, and \(\theta\) is the angle between the force and displacement.
Force Displacement Angle
The angle between the force applied and the direction of displacement has a significant impact on the work done on an object. This relationship is captured by the cosine term in the work done formula. \( \cos(\theta) \) varies from 1 to -1 as \( \theta \) varies from 0掳 to 180掳, affecting the amount of work done.

For instance, if the force is applied in the same direction as the displacement (\( \theta = 0掳 \) ), the cosine term is 1, and therefore, all of the force contributes to the work done. As in our exercise, the force was applied at a 25.0掳 angle to the displacement, meaning not all of the force's magnitude was effective in doing work on the block. For a force perpendicular to the displacement (\( \theta = 90掳 \) ), no work is done, which is demonstrated by both the normal force and gravitational force in the problem, where their work contribution is zero.
Normal and Gravitational Forces
Normal and gravitational forces are two fundamental forces that often act perpendicular to the displacement of objects on a horizontal surface. The normal force is the support force exerted by a surface perpendicular to the object, preventing it from falling through the surface. Gravitational force, commonly known as weight, is directed downwards towards the center of the Earth.

H4 sub-section: Work Done by Perpendicular Forces
It's important to realize that if either force acts at a 90掳 angle (perpendicular) to the displacement, the work done by that force will be zero, because \( \cos(90掳) = 0 \). This is exactly what happened in the problem with both the normal and gravitational forces. Their effect on the motion in the horizontal direction was nonexistent in terms of doing work, hence their work contributions were null.

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Most popular questions from this chapter

A crate of mass 10.0 kg is pulled up a rough incline with an initial speed of 1.50 m/s. The pulling force is 100 N parallel to the incline, which makes an angle of 20.0掳 with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 5.00 m. (a) How much work is done by the gravitational force on the crate? (b) Determine the increase in internal energy of the crate鈥搃ncline system due to friction. (c) How much work is done by the 100-N force on the crate? (d) What is the change in kinetic energy of the crate? (e) What is the speed of the crate after being pulled 5.00 m?

The direction of any vector \(A\) in three-dimensional space can be specified by giving the angles \(\alpha, \beta,\) and \(\gamma\) that the vector makes with the \(x, y,\) and \(z\) axes, respectively. If \(A=\) \(A_{x} \hat{i}+A_{y} \hat{j}+A_{z} \hat{k},\) (a) find expressions for \(\cos \alpha, \cos \beta,\) and \(\cos \gamma\) (these are known as direction cosines), and (b) show that these angles satisfy the relation \(\cos ^{2} \alpha+\cos ^{2} \beta+\) \(\cos ^{2} \gamma=1 .\) (Hint: Take the scalar product of \(\mathrm{A}\) with \(\hat{\mathrm{i}}, \hat{\mathrm{j}}\) , and \(\mathrm{k}\) separately.)

You can think of the work-kinetic energy theorem as a second theory of motion, parallel to Newton's laws in describing how outside influences affect the motion of an object. In this problem, solve parts (a) and (b) separately from parts (c) and (d) to compare the predictions of the two theories. In a rifle barrel, a 15.0 -g bullet is accelerated from rest to a speed of 780 \(\mathrm{m} / \mathrm{s}\) . (a) Find the work that is done on the bullet. (b) If the rifle barrel is 72.0 \(\mathrm{cm}\) long, find the magnitude of the average total force that acted on it, as \(F=W /(\Delta r \cos \theta) .\) (c) Find the constant acceleration of a bullet that starts from rest and gains a speed of 780 \(\mathrm{m} / \mathrm{s}\) over a distance of \(72.0 \mathrm{cm} .\) (d) If the bullet has mass 15.0 g, find the total force that acted on it as \(\Sigma F=m a\) .

Vector A has a magnitude of 5.00 units, and \(\mathbf{B}\) has a magnitude of 9.00 units. The two vectors make an angle of \(50.0^{\circ}\) with each other. Find \(\mathbf{A} \cdot \mathbf{B}\) .

While running, a person dissipates about 0.600 \(\mathrm{J}\) of mechanical energy per step per kilogram of body mass. If a 60.0 -kg runner dissipates a power of 70.0 \(\mathrm{W}\) during a race, how fast is the person running? Assume a running step is 1.50 \(\mathrm{m}\) long.
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