8.1 Energy (Page 2/3)

Siyavula textbooks: grade 12 Page 2 / 3

Capacity to do work

Energy is the capacity to do work. When positive work is done on an object, the system doing the work loses energy. In fact, the energy lost by a system is exactly equal to the work done by the system. An object with larger potential energy has a greater capacity to do work.

Show that a hammer of mass 2 kg does more work when dropped from a height of 10 m than when dropped from a height of 5 m. Confirm that the hammer has a greater potential energy at 10 m than at 5 m.

Determine what is given and what is required
We are given:
- the mass of the hammer, $m =$ 2 kg
- height 1, $h_{1}$ =10 m
- height 2, $h_{2}$ =5 m
We are required to show that the hammer does more work when dropped from $h_{1}$ than from $h_{2}$ . We are also required to confirm that the hammer has a greater potential energy at 10 m than at 5 m.
Determine how to approach the problem
1. Calculate the work done by the hammer, $W_{1}$ , when dropped from $h_{1}$ using:
  $W_{1} = F_{g} \cdot h_{1} .$
2. Calculate the work done by the hammer, $W_{2}$ , when dropped from $h_{2}$ using:
  $W_{2} = F_{g} \cdot h_{2} .$
3. Compare $W_{1}$ and $W_{2}$
4. Calculate potential energy at $h_{1}$ and $h_{2}$ and compare using:
  $P E = m \cdot g \cdot h .$
Calculate $W_{1}$
$\begin{matrix} W_{1} & = & F_{g} \cdot h_{1} \\ = & m \cdot g \cdot h_{1} \\ = & (2 kg) (9.8 m \cdot s^{- 2}) (10 m) \\ = & 196 J \end{matrix}$
Calculate $W_{2}$
$\begin{matrix} W_{2} & = & F_{g} \cdot h_{2} \\ = & m \cdot g \cdot h_{2} \\ = & (2 kg) (9.8 m \cdot s^{- 2}) (5 m) \\ = & 98 J \end{matrix}$
Compare $W_{1}$ and $W_{2}$
We have $W_{1}$ =196 J and $W_{2}$ =98 J. $W_{1} > W_{2}$ as required.
Calculate potential energy
From [link] , we see that:

$\begin{matrix} P E & = & m \cdot g \cdot h \\ = & F_{g} \cdot h \\ = & W \end{matrix}$

This means that the potential energy is equal to the work done. Therefore, $P E_{1} > P E_{2}$ , because $W_{1} > W_{2}$ .

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This leads us to the work-energy theorem.

Work-Energy Theorem

The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy:

W = Δ K E = K E_{f} - K E_{i}

The work-energy theorem is another example of the conservation of energy which you saw in Grade 10.

A brick of mass 1 kg is dropped from a height of 10 m. Calculate the work done on the brick at the point it hits the ground assuming that there is no air resistance?

Determine what is given and what is required
We are given:
- mass of the brick: $m$ =1 kg
- initial height of the brick: $h_{i}$ =10 m
- final height of the brick: $h_{f}$ =0 m
We are required to determine the work done on the brick as it hits the ground.
Determine how to approach the problem
The brick is falling freely, so energy is conserved. We know that the work done is equal to the difference in kinetic energy. The brick has no kinetic energy at the moment it is dropped, because it is stationary. When the brick hits the ground, all the brick's potential energy is converted to kinetic energy.
Determine the brick's potential energy at $h_{i}$
$\begin{matrix} P E & = & m \cdot g \cdot h \\ = & (1 kg) (9, 8 m \cdot s^{- 2}) (10 m) \\ = & 98 J \end{matrix}$
Determine the work done on the brick
The brick had 98 J of potential energy when it was released and 0 J of kinetic energy. When the brick hit the ground, it had 0 J of potential energy and 98 J of kinetic energy. Therefore $K E_{i}$ =0 J and $K E_{f}$ =98 J.

From the work-energy theorem:

$\begin{matrix} W & = & Δ K E \\ = & K E_{f} - K E_{i} \\ = & 98 J - 0 J \\ = & 98 J \end{matrix}$
Write the final answer
98 J of work was done on the brick.

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The driver of a 1 000 kg car traveling at a speed of 16,7 $m \cdot s^{- 1}$ applies the car's brakes when he sees a red robot. The car's brakes provide a frictional force of 8000 N. Determine the stopping distance of the car.

Determine what is given and what is required
We are given:
- mass of the car: $m$ =1 000 kg
- speed of the car: $v$ =16,7 $m \cdot s^{- 1}$
- frictional force of brakes: $F$ =8 000 N
We are required to determine the stopping distance of the car.
Determine how to approach the problem
We apply the work-energy theorem. We know that all the car's kinetic energy is lost to friction. Therefore, the change in the car's kinetic energy is equal to the work done by the frictional force of the car's brakes.

Therefore, we first need to determine the car's kinetic energy at the moment of braking using:

$K E = \frac{1}{2} m v^{2}$

This energy is equal to the work done by the brakes. We have the force applied by the brakes, and we can use:

$W = F \cdot d$

to determine the stopping distance.
Determine the kinetic energy of the car
$\begin{matrix} K E & = & \frac{1}{2} m v^{2} \\ = & \frac{1}{2} (1 000 k g) {(16, 7 m s)}^{2} \\ = & 139 445 J \end{matrix}$
Determine the work done
Assume the stopping distance is $d_{0}$ . Then the work done is:

$\begin{matrix} W = F \cdot d \\ = & (- 8 000 N) (d_{0}) \end{matrix}$

The force has a negative sign because it acts in a direction opposite to the direction of motion.
Apply the work-enemy theorem
The change in kinetic energy is equal to the work done.

$\begin{matrix} Δ K E & = & W \\ K E_{f} - K E_{i} & = & (- 8 000 N) (d_{0}) \\ 0 J - 139 445 J & = & (- 8 000 N) (d_{0}) \\ ∴ d_{0} & = & \frac{139 445 J}{8 000 N} \\ = & 17, 4 m \end{matrix}$
Write the final answer
The car stops in 17,4 m.

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Source: OpenStax, Siyavula textbooks: grade 12 physical science. OpenStax CNX. Aug 03, 2011 Download for free at http://cnx.org/content/col11244/1.2

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