Lesson Explainer: Work Done by a Constant Force | Nagwa Lesson Explainer: Work Done by a Constant Force | Nagwa

Lesson Explainer: Work Done by a Constant Force Mathematics

In this explainer, we will learn how to calculate the work done by a constant force acting on a particle.

Let us first define acceleration due to a force.

Definition: Acceleration due to a Force

When a net force acts on a body, the body accelerates in the direction of the force. The magnitude of the acceleration depends on the magnitude of the force and on the mass of the body according to the formula 𝐹=𝑚𝑎, where 𝑚 is the mass of the body and 𝑎 is the acceleration of the body.

The product of the force on the body and the displacement of the body parallel to the direction of the force while the force acts on it is equal to the work done on the body, 𝑊: 𝑊=𝐹𝑑.

The force that acts on a body does not necessarily act in the direction of the displacement of the body over the time interval that the force acts on the body. A body that is in motion may be acted on by a force that has a direction other than the direction of the velocity of the body.

This relationship allows the work done by a force to be defined.

Definition: Work Done on a Body by a Force

The work done on a body by a force is dependent on the force that acts on the body and the distance that the body moves in the direction of that force according to the formula 𝑊=𝐹𝑑𝜃,cos where 𝐹 is the magnitude of the force, 𝑑 is the magnitude of displacement of the body while the force acts on it, and 𝜃 is the angle between the directions of 𝐹 and 𝑑.

Let us look at an example of a force doing work on a body to change the kinetic energy of the body.

Example 1: Finding the Work Done on a Body given Its Acceleration and Distance Covered

A body of mass 0.9 kg moved a distance of 25 cm while accelerating at 8 cm/s2. Find the work done 𝑊.

Answer

The body accelerates. The 8 cm/s2 acceleration is converted to a 0.08 m/s2 acceleration so that the force is expressed in newtons. The acceleration and the mass of the body are known, so the force that acts on the body to accelerate it is then given by 𝐹=𝑚𝑎𝐹=0.9(0.08)=0.072.N

The body accelerates in the direction of the force, so the work done by the force is given by 𝑊=𝐹𝑑=0.072(0.25)=0.018.joules

The work can be expressed as an integer value in the unit ergs, where 10 ergs equal 1 joule: 𝑊=0.018×10=1.8×10.ergs

Let us now look at an example where a force does work on a body in the opposite direction to the velocity of the body.

Example 2: Finding the Work Done by a Constant Resistive Force

A body moves in a straight line against a resistance force of 360 N. Find the work done by this force over the displacement 𝑆, where 𝑆=390m.

Answer

The only force that acts on the body as it moves is the resistance force. A resistive force acting on a body must act in the opposite direction to the motion of the body, so the angle between the direction of motion and the force is 180 degrees. The work done by the resistive force is found using 𝑊=𝐹𝑑𝜃,cos where 𝜃=180.

So, cos𝜃=1.

The magnitude of the work done will therefore be multiplied by 1. The absolute magnitude will not be changed by this, but the work done will be negative. The work done by the resistive force is given by 360(390)(1)=140400.J

Let us look at an example of a force doing work on a body, where the velocity of the body reverses direction during the motion of the body.

Example 3: Finding the Work Done by an Upward Projected Body after a Given Interval of Time

A body of mass 900 g was projected vertically upward at 6.4 m/s. Find the work done by the weight of the body during the first 5 seconds of the motion of the body. Take 𝑔=9.8/ms.

Answer

In this example, a body has an initial velocity vertically upward. There is however no vertically upward force on the body. The only force on the body is the weight of the body. The weight of the body acts vertically downward with a constant magnitude given by 𝐹=0.9(9.8)=8.82.N

The 900 g mass of the object is converted to 0.9 kg so that the unit of the weight is newtons.

The body moves for at least 5 seconds. During that time the weight of the body acts on the body and so does work on the body.

To determine the work done on the body by its weight, we must determine the distance that the body moves in the direction that its weight acts: vertically downward. If the body moved vertically upward for 5 seconds, the displacement would be in the opposite direction to the direction that the weight acts, so the value of the work done depends on whether the net motion of the body is vertically upward or downward.

The net displacement of the body under constant acceleration 𝑔 is given by 𝑠=𝑢𝑡+12𝑎𝑡.

It is important to note that the initial velocity and the acceleration of the object are in opposite directions. The signs of 𝑢 and 𝑎 must be opposite. Taking 𝑎 as negative and substituting values from the question into the equation give 𝑠=6.4𝑡129.8𝑡𝑠=6.4(5)129.8(5)𝑠=90.5.m

The net displacement of the body is 90.5 m vertically downward.

Using 𝑊=𝐹𝑑, we can see that 𝑊=8.82(90.5)=798.21.J

Let us look at an example where a force does work on a body at an arbitrary angle to the velocity of the body.

Example 4: Finding the Work Done by an Inclined Force

How much work is done by a force of magnitude 200 newtons and direction 30 degrees from the horizontal that slides a crate 20 metres horizontally along a dock? Give your answer to two decimal places.

Answer

Perpendicular components of the 200 N force acting on the crate, where one of the components is parallel to the displacement of the crate, are shown in the following figure.

The work done by the 200 N force on the crate is given by 𝑊=𝐹𝑑𝜃𝑊=200(20)((30))𝑊=400032.coscos

Taking the value of 𝑊 to two decimal places gives 𝑊=3464.10,joules where 1=1×1.joulenewtonmetre

So, the result can be expressed as 𝑊=3464.10.newton-metres

Let us look at another example of a force acting at an arbitrary angle to the displacement of the body that the force acts on.

Example 5: Calculating the Amount of Work Done by a Force

The figure shows a force 𝐹 of magnitude 2 N and the displacement 𝑠=16m covered by a body acted on by the force. Find the work done by the force.

Answer

As the figure shows, the force acting on the body does not act in the direction of the displacement of the body. There is an angle between the direction of the force and the displacement that is not shown.

The angle from the line of action of 𝐹 is 180, as the line of action is a straight line. This angle consists of two angles, one of which is unknown and the other is shown in the figure to be 60. So, the unknown angle a is given by 𝑎=18060=120.

The formula for the work done on the body by the force is 𝑊=𝐹𝑑𝜃𝑊=2(16)((120))𝑊=3212=16.coscosJ

The work done by the force on the body is negative. The negative value of the work can be understood by consideration of the directions of the perpendicular components of 𝐹, where one of the components is parallel to 𝑠, as shown in the following figure.

The component of 𝐹 parallel to 𝑠 has the opposite direction to 𝑠.

Key Points

  • The work done on a body due to a force acting on it is the product of the force acting on it and the displacement of the body in the direction of the force according to the formula 𝑊=𝐹𝑑, where 𝐹 is the magnitude of the force and 𝑑 is the magnitude of the displacement of the body in the direction of the force.
  • The work done on a body by a force that does not act in the direction of the displacement of the body is equal to the work done by the component of the force that acts in the direction of the displacement of the body according to the formula 𝑊=𝐹𝑑𝜃,cos where 𝐹 is the magnitude of the force, 𝑑 is the magnitude of the displacement of the body while the force acts on it, and 𝜃 is the angle between the directions of 𝐹 and 𝑑.

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