Additional Answer Options: 5.3

2023 - May/June - Paper 1 - Question 5

5.1

 

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

light bulb Tip: a non-conservative force does not change the mechanical energy of a system.

2 marks

(1 x 2 marks)

5.2

 

Option 1

\left.\begin{array}{c}W_{\text{net }}=\Delta K\\ W_{w}+W_{f}+W_{F}=1/2mv_{f}^2-1/2mv_{i}^2\\ mg\sin\theta\Delta x\cos\theta+W_{f}+W_{F}=1/2mv_{f}^2-1/2mv_{i}^2\end{array}\right\}Any one (1)check

(20)(9,8)\left(\sin18^{\circ}\right)(15,6)\cos180^{\circ}(13,5)(15,6)\cos180^{\circ}+(96,8)(15,6)\cos0^{\circ}

=\frac12(20)\left(v_{f}^2-0^2\right)(2 + 3 + 4) checkcheckcheck

v_{f}=5,96\mathrm{~m}\cdot\mathrm{s}^{-1} (5)check

light bulb Tips: 

  • The crate starts from rest and accelerates up the slope.
  • It helps to draw a free-body diagram to show all of the forces acting on an object in order to identify which of the forces do the work.

  • In this case, negative work is done against gravity and friction. Positive work is done by the tension in the rope. Using the formula W = FΔxcosθ, calculate the work done by each of the forces acting on the crate.
  • The work done by the frictional force is equal to Wf = (13,5)(15,6)cos 180°
  • The work done against gravity is equal to Wg = (20)(9,8)(15,6sin18°)cos180°
  • The work done by the tension is equal to WT = (96,8)(15,6)cos0°
  • The normal force does no work since the normal force acts perpendicularly to the direction of the displacement.
  • Using the work-energy theorem Wnet = ΔEk

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

5 marks

(5 x 1 mark)

Marking criteria:

  • (1) Any one of the correct equation check
  • (2) Correct substitution for work done by gravity or ΔEp check
  • (3) Correct substitution for work done by motor and friction check
  • (4) Correct substitution for ∆Ek check
  • (5) Correct final answer: 5,96 m·s-1 check 

5.3

 

Option 1

\mathrm{P}_{\mathrm{ave}}=\mathrm{Fv}_{\mathrm{ave}} (1)check

=96,8\left(\frac{(0)+5,96}{2}\right) (2)check

=288,46\mathrm{~W} (3)check

light bulb Tips: 

  • The average power is the product of the applied force of the motor and the average velocity of the crate.
  • Since the crate accelerates uniformly from rest to a speed of 5,96 m·s-1, the average speed is the average of these two velocities.

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

3 marks

(3 x 1 mark)

Marking criteria:

  • (1) Correct equation for power check
  • (2) Correct substitution into power equation check
  • (3) Correct final answer: 288,46 Wcheck
  • Range: 286,46 W to 288,73 W

5.4

 

checkcheckcheck

light bulb  Tips: 

  • Although older papers may indicate that you may also use components in free-body diagrams, they will no longer be marked be awarded the marks.
  • Do not resolve the weight into parallel and perpendicular components in any free-body diagram. 
Ncheck FN/196N/normal/Fnormal
fcheck (kinetic) friction/Ff/Fw/fk
wcheck Fg/Fw/weight/mg/gravitational force/FEarth on crate

3 marks

(3 x 1 mark)

5.5

 

light bulb How to approach this question: 

  • If upwards is taken as the positive direction.

checkcheckcheckcheck

  • When the rope breaks, the crate continues to move up the slope but slows down until it comes to rest at point D. Then, it slides down the slope with a less steep gradient because friction and gravity act in opposite directions.

Or

  • If upwards is considered as negative.

checkcheckcheckcheck

4 marks

(4 x 1 mark)

Marking criteria:

  • The first straight line starts at zero with a positive gradient, reaching a maximum
    velocity. check
  • A second straight line with the negative gradient from maximum velocity to zero. check
  • The third straight line continues from the second line at zero and extends below the x-axis. check
  • The third line has a smaller negative gradient than the second line. check

Or

Part 1

Option 1

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

15,6=\left(\frac{(0)+5,96}{2}\right)\Delta t

\Delta\mathrm{t}=5,23\mathrm{~s}(5,24)

Or

Option 2

v_{f}^2=v_{i}^2+2a\Delta x

(5,96)^2=0^2+2a(15,6)

\mathrm{a}=1,14\mathrm{~m}\cdot\mathrm{s}^{-2}\quad(1,13)

v_{\mathrm{f}}=\mathrm{v}_{\mathrm{i}}+\mathrm{a}\Delta\mathrm{t}

5,96=0+(1,14)\Delta t

\Delta\mathrm{t}=5,23\mathrm{~s}\quad(5,27)

Part 2

Option 1

\left.\begin{array}{l}P=\frac{W}{\Delta t}\\ P=\frac{F\Delta x\cos\theta}{\Delta t}\end{array}\right\}Anyone (1)

P=\frac{96,8(15,6)\cos0}{5,23} (2)

P=288,73\mathrm{~W} (3)

Or

Option 2:

\mathrm{v}_{\mathrm{ave}}=\frac{\Delta\mathrm{x}}{\Delta\mathrm{t}}

=\frac{15,6-0}{5,23-0}

=2,98\mathrm{~m}\cdot\mathrm{s}^{-1}

\mathrm{P}_{\mathrm{ave}}=\mathrm{Fv}_{\mathrm{ave}} (1)

=96,8(2,98) (2)

=288,46\mathrm{~W} (3)

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