Additional Answer Options: 5.5

2018 - May/June - Paper 1 - Question 5

5.1

 

 checkcheck

Or

 checkcheck  

light bulb Tips: 

  • A free-body diagram shows all of the forces acting on the object.
  • In this case, the only forces acting on the object are the weight downwards and the tension upwards.

2 marks

(2 x 1 mark)

5.2

Tension check

light bulb Tips: 

  • The work done by a non-conservative force is dependent on the path taken.
  • Gravitational force is a conservative force and tension is a non-conservative force.

1 mark

5.3

Option 1

\left.\begin{array}{l}\mathrm{W}=\mathrm{F}\Delta\mathrm{x}\cos\theta\\ \mathrm{W}_{\mathrm{w}}=\mathrm{mg}\Delta\mathrm{x}\cos\theta\end{array}\right\} Any one check 

=75(9,8)(12) \cos 180^{\circ} check 

=-8820 \mathrm{~J} check 

light bulb How to approach this question: 

  • The work done by the gravitational force can be calculated using W = FΔxcosθ, the angle θ is 180° because the weight acts downwards but the load moves upwards.

Or

Option 2

\mathrm{W}_{\mathrm{w}}=-\Delta \mathrm{E}_{\mathrm{p}} check 

=-(m g h-0) 

=-(75)(9,8)(12) check 

=-8820 \mathrm{~J} check 

3 marks

(3 x 1 mark)

5.4

 

The work done on an object by a net force is equal to the change in the object’s kinetic energy. checkcheck

Or

The net work done on an object is equal to the change in the object’s kinetic energy. checkcheck

2 marks

(1 x 2 marks)

5.5

 

light bulb How to approach this question: 

  • The instructions specifically instruct you to use the work-energy theorem, therefore, you will not be awarded marks for using another method even if you get the same answer.

Option 1

\left.\begin{array}{l}\mathrm{W}_{\text{net }}=\Delta\mathrm{K}\\ \mathrm{F}_{\mathrm{net}}\Delta\mathrm{x}\cos\theta=\left(\frac12\mathrm{mv}_{\mathrm{f}}^2-\frac12\mathrm{mv}_{\mathrm{i}}^2\right)\end{array}\right\} Any one check 

(75)(0,65)(12)\cos0^{\circ}-=\frac12(75)\left(v_{f}^2-0\right) checkcheckcheck 

\mathrm{v}_{\mathrm{f}}=3,95 \mathrm{~m} \cdot \mathrm{s}^{-1}\left(3,949 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) check 

  • Fnet = ma in this case, Fnet = (75)(0,65).
  • Often you would work out the Fnet by adding together the "work" done by a number of different forces; an alternative is to use the resultant force.

Or

Option 2

\left.\begin{array}{l}\mathrm{W}_{\text{net }}=\Delta\mathrm{K}\\ \mathrm{W}_{\mathrm{nc}/nk}=\Delta\mathrm{K}+\Delta\mathrm{U}\\ \mathrm{W}_{\mathrm{T}}+\mathrm{W}_{\mathrm{g}}=\Delta\mathrm{K}\end{array}\right\} Any one check 

\mathrm{T}-\mathrm{mg}=\mathrm{ma} 

T-75(9,8)=75(0,65) check 

\mathrm{T}=783,75 \mathrm{~N} 

\mathrm{W}_{\mathrm{T}}=783,75(12) \cos 0^{\circ} check 

=9405 \mathrm{~J} 

9405-(8820)=1 / 2(75)\left(v_f^2-0\right) check 

v_{\mathrm{f}}=3,95 \mathrm{~m} \cdot \mathrm{s}^{-1}\left(3,949 \mathrm{~m} \cdot \mathrm{s}^{-1}\right) check 

There are more ways to answer this question. To view other options, click here.

5 marks

(5 x 1 mark)

Or

Option 2

\left.\begin{array}{l}W_{nc}=\Delta E_{p}+\Delta E_{k}\\ f\Delta x\cos\theta=\left(mgh_{f}-mgh_{i}\right)+\left(1/2mv_{f}^2-1/2mv_{i}^2\right)\end{array}\right\} Any one

\mu mg\Delta x\cos180^{\circ}=\left[0-\left(mg\Delta x\sin30^{\circ}\right]+1/2m\left(v_{f}^2-v_{i}^2\right)\right.

[33,948)(5)(-1)]=\left[0-50(9,8)(5)\sin30^{\circ}\right]+1/2(50)\left(v_{f}^2-7,668^2\right)

v_{\mathrm{f}}=10,05\mathrm{~m}\cdot\mathrm{s}^{-1}

Or

Option 3

\left.\begin{array}{l}W_{nc}=\Delta E_{p}+\Delta E_{k}\\ f\Delta x\cos\theta=\left(mgh_{f}-mgh_{i}\right)+\left(1/2mv_{f}^2-1/2mv_{i}^2\right)\end{array}\right\} Any one

(33,95)(5)\cos180=\left[(50)(9,8)\left\{0-5\sin30^{\circ}\right\}\right]+1/2(50)\left(v_{f}^2-7,668^2\right)

\mathrm{v}_{\mathrm{f}}=10,05\mathrm{~m}\cdot\mathrm{s}^{-1}

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