Additional Answer Options: 5.5

2018 - November - Paper 1 - Question 5

5.1

A force is non-conservative if the work it does on an object which is moving between two points depends on the path taken.checkcheck

Or

A force is non-conservative if the work it does on an object depends on the path taken.checkcheck

Or

A force is non-conservative if the work it does in moving an object around a closed path is non-zero. checkcheck

2 marks

(1 x 2 marks)

5.2

No check

light bulbTip: force F is the force of the engine; it increases the mechanical energy of the system, therefore, it is a non-conservative force.

1 mark

5.3

Option 1

P=\frac{W}{\Delta t}check

=\frac{4,8\times10^6}{(90)}check

=53333,33\mathrm{~W}

=5,33\times10^4\mathrm{~W}(53,33\mathrm{~kW})check

Or

Option 2

\Delta\mathrm{x}=\left(\frac{\mathrm{v}_{\mathrm{f}}+\mathrm{v}_{\mathrm{i}}}{2}\right)\Delta\mathrm{t}

=\left(\frac{0+25}{2}\right)(90)

=1125\mathrm{~m}

W_{F}=F\Delta x\cos\theta

4,80\times10^6=F(1125)\cos0^{\circ}

F=4266,667N

\mathrm{P}_{\mathrm{ave}}=\mathrm{Fv}_{\mathrm{ave}}check

=(4266,667)(12,5)check

=53333,33\mathrm{~W}check

light bulb Tips: 

  • Power is the rate at which work is done.
  • Many different equations can be used to calculate the power, but the first option is the most direct one since the work done and time taken are both given.
  • The second option involves many more steps, however, you can still get full marks if you do all of the steps.

3 marks

(3 x 1 mark)

5.4 

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

Or

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

light bulb  Tip: the work-energy theorem is asked frequently in exams. Ensure that you know the definition well, do not wait until right before the exam to memorise it.

2 marks

(1 x 2 marks)

5.5

 

light bulb Tips: 

  • This formula comes directly off the formula sheet, and it should be written exactly as it is. Alternatively, you can use Wnet = ΔEk

Option 1

\mathrm{W}_{\text{net }}=\Delta\mathrm{K}check

W_{W}+W_{f}+W_{F}=1/2mv_{f}^2-1/2mv_{i}^2

mg\Delta x\cos\theta+W_{f}+W_{F}=1/2mv_{f}^2-1/2mv_{i}^2

(1500)(9,8)200\cos180^{\circ}+W_{f}+4.8\times10^6=1/2(1500)\left(25^2-0\right)checkcheckcheck

  • Three different forces are acting on the object; the force of gravity, friction and the applied force of the engine. All three of these forces do work on the object.

-2940000+W_{f}+4,8\times10^6=468750

W_{f}=-1391250\mathrm{~J}

=-1,39\times10^6\mathrm{~J}check

  • The work done by friction is negative. Work done by friction is always negative since friction always acts in the opposite direction to the direction of motion.
  • It is really important that you use an energy principle, if you use any other method, you will not be awarded any marks even if you get the correct answer.

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}{rl}\mathrm{W}_{\mathrm{nc}} & =\Delta\mathrm{K}+\Delta\mathrm{U}\\ \mathrm{W}_{\mathrm{nc}} & =1/2mv_{\mathrm{f}}{}^2-1/2mv_{\mathrm{i}}{}^2+\mathrm{mgh}_{\mathrm{f}}-\mathrm{mgh}_{\mathrm{i}}\\ & =1/2m\left(\mathrm{v}_{\mathrm{f}}{}^2-v_{\mathrm{i}}{}^2\right)+mg\left(h_{\mathrm{f}}-\mathrm{h}_{\mathrm{i}}\right)\\ \mathrm{W}_{\mathrm{nc}} & =1/2mv_{\mathrm{f}}{}^2+mgh_{\mathrm{f}}-1/2mv_{\mathrm{i}}{}^2-mgh_{i}\end{array}\right\}Any one

W_{f}+W_{F}=1/2mv_{f}^2-1/2mv_{i}^2+mqh_{f}-mah_{i}

W_{f}+4.8\times10^6=\left[1/2(1500)(25)^2+-0\right]+[(1500)(9,8)(200)-0]

W_{f}=-1,39\times10^6\mathrm{~J}\left(-1,40\times10^6\mathrm{~J}\right)

Or

Option 3

\left.\begin{array}{l}W_{nc}=\Delta K+\Delta U\\ W_{nc}=1/2mv_{f}^2-1/2mv_{i}^2+mgh_{f}-mgh_{i}\\ \quad=1/2m\left(v_{f}^2-v_{i}^2\right)+mg\left(h_{f}-h_{i}\right)\\ W_{nc}=1/2mv_{f}^2+mgh_{f}-1/2mv_{i}^2-mgh_{i}\end{array}\right\}Any one

W_{f}+4.8\times10^6=\left[1/2(1500)(25)^2+(1500)(9.8)(200)\right]-[0+0]

W_{f}=-4,8\times10^6+3,4\times10^6

=-1,39\times10^6\mathrm{~J}\left(-1,40\times10^6\mathrm{~J}\right)

Related subjects & topics
Physical Sciences
Explore similar posts in our community