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A 70 kg box is initially at rest at the bottom of a ROUGH plane inclined at an angle of 30° to the horizontal - NSC Physical Sciences - Question 5 - 2019 - Paper 1

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A 70 kg box is initially at rest at the bottom of a ROUGH plane inclined at an angle of 30° to the horizontal. The box is pulled up the plane by means of a light ine... show full transcript

Worked Solution & Example Answer:A 70 kg box is initially at rest at the bottom of a ROUGH plane inclined at an angle of 30° to the horizontal - NSC Physical Sciences - Question 5 - 2019 - Paper 1

Step 1

5.1 What is the name given to the force in the rope?

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Answer

The force in the rope is referred to as 'Tension' or 'Spanning'.

Step 2

5.2 Give a reason why the mechanical energy of the system will NOT be conserved as the box is pulled up the plane.

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Answer

The mechanical energy of the system will not be conserved due to the presence of friction, which is a non-conservative force. Energy is lost to heat through friction as the box moves up the incline.

Step 3

5.3 Draw a labelled free-body diagram for the box as it moves up the plane.

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Answer

The free-body diagram includes:

  • Weight (W) acting downwards, equal to mg.
  • Tension (T) acting parallel to the incline.
  • Normal force (N) acting perpendicular to the incline.
  • Frictional force (f) acting in the opposite direction to the movement.

The diagram should show the forces clearly with appropriate labels.

Step 4

5.4 Calculate the work done on the box by the frictional force over the 4 m.

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Answer

The work done by the frictional force can be calculated using the formula:

W=Fimesdimesextcos(180°)W = F imes d imes ext{cos}(180°)

Where:

  • F=178.22NF = 178.22 \, N (frictional force)
  • d=4md = 4 \, m (distance moved)

Thus, W=178.22imes4imes(1)=712.88JW = 178.22 imes 4 imes (-1) = -712.88 \, J

The negative sign indicates that work is done against the direction of movement.

Step 5

5.5 Use energy principles to calculate the speed of the box after it has moved 4 m.

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Answer

To find the speed after moving 4 m, we can apply the work-energy principle:

  • The net work done on the box is equal to the change in kinetic energy:

Wnet=extKEfinalextKEinitialW_{net} = ext{KE}_{final} - ext{KE}_{initial}

Since the box starts from rest: extKEinitial=0 ext{KE}_{initial} = 0

Thus, W_{net} = rac{1}{2} mv^2

Given the net work done is from the applied force and friction: Wnet=(700imes4)712.88W_{net} = (700 imes 4) - 712.88 extSolvingforthespeed,v: ext{Solving for the speed, } v: After calculations, we find that: v=4.52m/sv = 4.52 \, m/s

Step 6

5.6 What will be the total work done by friction when the box moves up and then to the bottom of the inclined plane?

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Answer

The total work done by friction consists of two phases:

  1. Work done against friction as the box moves up:
    • Wfriction,up=712.88JW_{friction, up} = -712.88 \, J (calculated above)
  2. Work done by friction as the box slides back down:
    • The box will lose the same amount of energy as it gains going up, but since it is moving down, it is still in the direction against friction:
    • Wfriction,down=712.88JW_{friction, down} = 712.88 \, J (the same magnitude, opposite direction)

Total work done by friction during both motions is thus: TotalWfriction=Wfriction,up+Wfriction,down=712.88+712.88=0Total \, W_{friction} = W_{friction, up} + W_{friction, down} = -712.88 + 712.88 = 0

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