GCSE Physics (9-1)

2.3 Work done and energy transfer — Practise with help

Work each problem first. Open a hint only if you are stuck, and the full working only to check.

Warm-up: remember

  1. A spring of spring constant 200 N/m is stretched by 0.10 m. Calculate the force applied.
  2. Name the store that fills when work is done against friction.
Answers
  1. F=ke=200×0.10=20 NF = ke = 200 \times 0.10 = 20\,\text{N}.
  2. The thermal store of the surfaces.

If question 1 went wrong, return to Section 2.2 and the equation F=keF = ke.

Practise with help

1. A force of 90 N pushes a crate 6.0 m along the line of the force. Calculate the work done.

Hint

Work done needs the force and the distance moved along the line of the force.

Next step

Write W=FsW = Fs, then multiply 90 N by 6.0 m.

Full working W=Fs=90×6.0=540 J\begin{aligned} W &= F s \\ &= 90 \times 6.0 \\ &= 540\,\text{J} \end{aligned}

2. A machine does 1200 J of work over a distance of 25 m. Calculate the force.

Hint

You need the force on its own, so rearrange W=FsW = Fs.

Next step

Divide the work done by the distance: F=WsF = \dfrac{W}{s}.

Full working F=Ws=120025=48 N\begin{aligned} F &= \frac{W}{s} \\ &= \frac{1200}{25} \\ &= 48\,\text{N} \end{aligned}

3. A bag of mass 6.0 kg is lifted 1.5 m onto a wall. Take g=9.8 N/kgg = 9.8\,\text{N/kg}. Calculate the work done against gravity.

Hint

The force that does the work is the weight of the bag, not the mass.

Next step

Work out the weight first: F=mg=6.0×9.8F = mg = 6.0 \times 9.8. Then multiply by the height.

Full working F=mg=6.0×9.8=58.8 NW=Fs=58.8×1.5=88.2 J=88 J(2 significant figures)\begin{aligned} F &= m g = 6.0 \times 9.8 = 58.8\,\text{N} \\ W &= F s = 58.8 \times 1.5 = 88.2\,\text{J} \\ &= 88\,\text{J} \quad (2\ \text{significant figures}) \end{aligned}

4. A box slides across a rough floor and slows to a stop. Describe the energy transfer that takes place, and name the store that receives the energy.

Hint

Think about the force between the box and the floor as the box slides.

Next step

Name the store that decreases and the store that increases, and say what causes the transfer.

Full working

The kinetic store of the box decreases as the box slows. The friction force between the box and the floor does work, and the energy is transferred to the thermal store of the box and the floor. Their temperature rises. This answer is a model answer for you to compare with your own wording.

5. A bag of mass 20 kg is lifted 1.8 m onto a shelf. Take g=9.8 N/kgg = 9.8\,\text{N/kg}. Say which force does the work, then calculate the work done.

Hint

The lifting force must be at least as large as the weight, and it acts through the height lifted.

Next step

Find the weight with W=mgW = mg, then use W=FsW = Fs with the height as the distance.

Full working

The lifting force does the work against gravity. Its size is

F=mg=20×9.8=196 NF = m g = 20 \times 9.8 = 196\,\text{N}

Then

W=Fs=196×1.8=352.8 J=350 J(2 significant figures)\begin{aligned} W &= F s \\ &= 196 \times 1.8 \\ &= 352.8\,\text{J} \\ &= 350\,\text{J} \quad (2\ \text{significant figures}) \end{aligned}

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