# Work

Show/Hide Sub-topics (Work, Energy & Power | O Level Physics)
Show/Hide Sub-topics (Work, Energy and Power | A Level Physics)

## Work

Work is done when a force moves the object in the direction of the force and is given by the product of the force and the distance moved in the direction.

• $W = F \times d$, Where W= Work, F = Constant Force (N), d = Distance moved in the direction of the force (m)
• SI Unit for work is joules (J)
• One joule is defined as the work done when a force of one newton (N) moves an object through a distance of  one metre (m) in the direction of the force
• Other unit for work: Nm (Newton metre), which is obtained by multiplying the unit for force and distance.

## Conditions When NO Work Is Done

NO work is done when

1. the object does not move (E.g. A boy pushing against a tree does no work)
2. the direction of the force and the direction in which the point of application moves are perpendicular to one another (E.g. A boy carrying a stack of books while walking. No work is done on the stack of books in the upward direction as the stack of book is only moving horizontally.)

If no work is done for the 2 scenarios above, WHY do you feel tired (if you are the one doing the activity)?

For Scenario A:

When your muscles are used to exert a force on something, the individual muscle fibers are in a constant process of contracting and releasing to exert a steady force on an external object. The contracting and releasing in those muscle fibers involves force and motion, and hence, they are considered as internal work in your body. The energy shows up as warming in your muscle tissue, but if the object (tree) doesn’t move, no work is done on the object (tree).

For Scenario B:

If the box is being carried at constant velocity, then no net force is necessary to keep it in motion. The force exerted by the person is an upward force equal to the weight of the box, and that force is perpendicular to the motion. If there is no motion in the direction of the force, then no work in done by that force. However, you certainly feel like you are doing work if you carry a heavy box. The resolution of the paradox is similar to scenario A – your muscles must maintain an extra tension to stay upright under the load. This requires a greater amount of internal contraction and release of our muscle fibers, and hence internal work in our bodies. But the work done on the box is zero since by moving in a straight line at constant speed, the force (which you exert on the object) does NOT move the object in the direction of the force.

## Work To Kinetic Energy

When a force moves an object, it does work on the object and the object gains kinetic energy. If there are no dissipative forces (i.e. friction/air resistance) on an object (e.g. a box), 5 joules of work done on pushing the object will translate to the object gaining 5 joules of kinetic energy.

## Work To Gravitational Potential Energy

Recall that gravitational potential energy is the energy a body has due to its position above a reference level (i.e. ground). In order to find the gravitational potential energy of an object near the surface of the Earth, we need to consider the work done to lift the object of mass $m$ to a height $h$ above the reference level at constant speed.

Consider an object with mass $m$ at ground level. A force $F$ is exerted on the object in the upward direction in order to lift the object. The object is lifted at constant speed to a height of $h$ above the ground level.

Since the object is lifted at constant speed, the force $F$ is just the weight of the object. (i.e. $F = mg$) Why? If the force is larger than the weight of the object, there will be a net resultant force and the object will accelerate according to Newton’s Second Law. If the force is smaller than the weight of the object, the force will not be able to lift the object.

Since the work done by the object is given by:

$$W = F \times h$$

We will obtain:

\begin{aligned} W &= mg \times h \\ &= mgh \end{aligned}

From the above, we “derived” the equation for gravitational potential energy from work done equation!

## Worked Examples

### Example 1

A librarian applies a force $F = 8 \, \text{N}$ to push a book trolley, causing it to move a distance of $d = 5 \, \text{m}$ in the direction of the force. Calculate the work done on the trolley. Additionally, describe the primary energy transfer process when the librarian is pushing the trolley.

To calculate the work done $W$, the formula $W = F \times d$ is used:

$$W = 8 \, \text{N} \times 5 \, \text{m} = 40 \, \text{J}$$

The main energy transfer process when the librarian pushes the trolley is the conversion of mechanical work into kinetic energy. The librarian’s exerted force does work on the trolley, resulting in the trolley gaining kinetic energy as it moves. This process illustrates the transformation of external mechanical energy into the internal kinetic energy of the trolley.

### Example 2: Pushing A Box

A boy pushes a box across a rough horizontal floor. He exerts 5 N to move it by 2 m. What is the work done by the boy?

\begin{aligned} \text{Work done by boy} &= f \times d \\ &= 5 \, \text{N} \times 2 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

### Example 3: Pulling A Wagon

A boy pulls a 1 kg toy wagon along a smooth (i.e. no friction) horizontal floor over a distance of 5 m. If the speed of the wagon increases at a constant rate of $2 \, \text{m s}^{-2}$, what is the work done by the boy?

The net resultant force on the toy wagon can be given by Newton’s Second Law:

$$f_{\text{resultant}} = ma$$

Since the floor is smooth (no friction), the net resultant force on the toy wagon is just the force exerted by the boy.

Hence,

\begin{aligned} \text{Net Resultant Force} &= \text{Force exerted by boy} \\ \text{Force exerted by boy} &= m \times a \\ &= 1 \, \text{kg} \times 2 \, \text{m s}^{-2} \\ &= 2 \, \text{N} \end{aligned}

The work done by the boy will just be:

\begin{aligned} \text{Work done by boy on wagon} &= \text{force} \times \text{dist. moved in direction of force} \\ &= 2 \, \text{N} \times 5 \, \text{m} \\ &= 10 \, \text{J} \end{aligned}

### Example 4: Raising A Box

A 5 kg box is raised 50 m from its original position. What is the gain in gravitational potential energy? Assume gravitational field strength = $10 \, \text{N kg}^{-1}$. How much work is needed to raise the box to its new position?

Note: Assume that no energy is lost to air resistance.

\begin{aligned} E_{p} &= mgh \\ &= 5 \, \text{kg} \times 10 \, \text{N kg}^{-1} \times 50 \, \text{m} \\ &= 2500 \, \text{J} \end{aligned}

### Example 5: Dropping A Ball

A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level).

1. What is it’s initial gravitational potential energy?
2. Determine its velocity just before it hits the ground.

Note: Assume that no energy is lost to air resistance.

Part a:

Taking the ground level to be the reference level,

(For gravitational potential energy, the relevant formula is:)

\begin{aligned} E_{p} &= mgh \\ &= \left( 0.5 \, \text{kg} \right) \left( 10 \, \text{N kg}^{-1} \right) \left( 10 \, \text{m} \right) \\ &= 50 \, \text{J} \end{aligned}

Part b:

From the Principle of Conservation of Energy, the initial gravitational potential energy will be converted to its final kinetic energy. (I.e. Gain in kinetic energy = loss in gravitational potential energy)

For kinetic energy, the relevant formula is:

\begin{aligned} E_{k} &= \frac{1}{2} m v^{2} \\ 50 \, \text{J} &= \frac{1}{2} \left( 0.5 \, \text{kg} \right) v^{2} \\ v &= 14.1 \, \text{m s}^{-1} \end{aligned}

### Example 6: Dropping A Ball With The Effects of Air Resistance

A ball with a mass of 500 g is dropped from a height of 10 m from the ground level (reference level). There is an energy loss of $10 \, \text{J}$ due to air resistance. Determine its velocity just before it hits the ground.

You can obtain the gravitational potential energy of the ball from Example 5, which is $50 \, \text{J}$.

By the Principle of Conservation of Energy, the initial gravitational potential energy will be converted to final kinetic energy and energy lost due to air resistance. Hence, we have:

\begin{aligned} E_{\text{initial}} &= E_{\text{final kinetic energy}} +\text{Energy Lost} \\ 50 \, \text{J} &= E_{\text{final kinetic energy}} + 10 \, \text{J} \\ E_{\text{final kinetic energy}} &= 50 \, \text{J}-10 \, \text{J} \\ &= 40 \, \text{J} \end{aligned}

Using the formula for kinetic energy, the velocity is:

\begin{aligned} E_{\text{final kinetic energy}} &= \frac{1}{2}mv^{2} \\ 40 \, \text{J} &= \frac{1}{2} \left( 0.5 \text{kg} \right) v^{2} \\ v &= 12.6 \, \text{m s}^{-1} \end{aligned}

### Example 7: The Ball Rebounds

A ball with a mass of 500 g is thrown vertically downwards from a height of 10 m with a velocity of $5.0 \, \text{m s}^{-1}$. It hits the ground and bounces. It then rises to a maximum height of 8.0 m. Determine the energy loss due to the bounce.

Assume that no energy is lost due to air resistance.

From the Principle of Conservation of Energy, the initial kinetic energy and initial gravitational potential energy is equals to the final kinetic energy + final gravitational potential energy + energy lost due to bounce. Let’s put this into an equation form:

\begin{aligned} E_{\text{k, initial}} + E_{\text{p, initial}} &= E_{\text{k, final}} + E_{\text{p, final}} + \text{Energy lost} \\ \frac{1}{2}mv^{2} + mgh_{\text{initial}} &= 0 + mgh_{\text{final}} + \text{Energy lost} \\ \frac{1}{2}\left( 0.5 \right) \left( 5^{2} \right) + \left( 0.5 \right) \left( 10 \right) \left( 10 \right) &= \left( 0.5 \right) \left( 10 \right) \left( 8 \right) + \text{Energy lost} \\ 56.25 \, \text{J} &= 40 \, \text{J} + \text{Energy lost} \\ \text{Energy lost} &= 16.25 \, \text{J} \\ \text{Energy lost} &= 16.3 \, \text{J} \, \left(3 \, \text{s.f.} \right) \end{aligned}

### Example 8: Work Done By Car Engine

The engine of a car exerts a constant force of 10 kN. The car is able to accelerate constantly from rest to a speed of $30 \, \text{m s}^{-1}$ in 10 seconds. Determine the work done by the engine of the car and the kinetic energy of the car at the end of the 10 seconds.

From what you have learnt in Kinematics,

\begin{aligned} v &= u + at \\ a &= \frac{\left( 30-0 \right)}{10} \\ &= 3 \, \text{m s}^{-1} \end{aligned}

With the given engine force and calculated acceleration, we can find the mass of car by employing the use of Newton’s Second Law. (Note: The engine force is the net resultant force acting on the car. Can you see why?)

\begin{aligned} F &= ma \\ m &= \frac{F}{a} \\ &= \frac{10000}{3} \\ &= 3330 \, \text{kg} \end{aligned}

The distance travelled by the car in 10 seconds is given by:

\begin{aligned} d &= \frac{1}{2} \left( v + u \right) \times t \\ &= 150 \, \text{m} \end{aligned}

The work done by the engine of the car will be:

\begin{aligned} W &= F \times d \\ &= 10000 \, \text{N} \times 150 \, \text{m} \\ &= 1 500 000 \, \text{J} \\ &= 1.5 \times 10^{6} \, \text{J} \end{aligned}

The kinetic energy of the car at the end of 10 seconds will be given by:

\begin{aligned} E_{k} &= \frac{1}{2} mv^{2} \\ &= \frac{1}{2} \left( 3330 \, \text{kg} \right) \left( 30 \, \text{m s}^{-1} \right)^{2} \\ &= 1.5 \times 10^{6} \end{aligned}

We notice that the work done by the engine of the car is the same as the kinetic energy of the car. This shows that all the work done by the engine of the car is converted into the kinetic energy of the car. (which is true if there is no dissipative forces acting on the car)

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