5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q. Points P and Q on the ramp are 25 m apart. The skateboard passes point P at a speed $v_i$ and passes point Q at a speed of 15 m/s. Ignore rotational effects due to the wheels of the skateboard.

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5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q - NSC Physical Sciences - Question 5 - 2017 - Paper 1

Question 5

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5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q. Points P and Q on the ramp a... show full transcript

Worked Solution & Example Answer:5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q - NSC Physical Sciences - Question 5 - 2017 - Paper 1

Step 1

Draw a labelled free-body diagram

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Answer

To draw the free-body diagram:

Identify Forces:
- Gravitational Force ( $F_g$ ): Points downwards and is calculated as $F_g = mg$ , where $m = 60 ext{ kg}$ and $g = 9.8 ext{ m/s}^2$ . This results in a force of $588 ext{ N}$ .
- Normal Force ( $N$ ): Acts perpendicular to the ramp's surface and is influenced by the ramp's angle of $20^{ ext{o}}$ .
- Frictional Force ( $F_f$ ): Acts parallel to the ramp up the incline, given as $50 ext{ N}$ .
Components of Gravitational Force: Resolve $F_g$ into two components: one parallel to the incline and one perpendicular.
- Parallel component: $F_{ ext{parallel}} = mg imes ext{sin}(20^{ ext{o}})$
- Perpendicular component: $F_{ ext{perpendicular}} = mg imes ext{cos}(20^{ ext{o}})$
Label All Forces: Clearly label each force with arrows indicating direction. The gravitational force will act downwards while the normal force acts upwards, perpendicular to the inclined surface, and the frictional force opposes the motion.

Step 2

State the work-energy theorem in words.

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Answer

The work-energy theorem states that the net work done on an object is equal to the change in the object's kinetic energy. In mathematical terms, it can be expressed as: $W_{ ext{net}} = rac{1}{2} m v_f^2 - rac{1}{2} m v_i^2$

Step 3

Use energy principles to calculate the speed $v_i$ of the skateboarder at point P.

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To find the initial speed $v_i$ , we apply the work-energy theorem:

The net work done ( $W_{ ext{net}}$ ) is equal to the change in kinetic energy: $W_{ ext{net}} = F_{ ext{f}} imes d imes ext{cos}( heta)$ where $heta = 180^{ ext{o}}$ since friction acts opposite the motion. $W_{ ext{net}} = (50 ext{ N})(25 ext{ m})(-1) = -1250 ext{ J}$
Setting the work done equal to the change in kinetic energy: $-1250 = rac{1}{2}(60)(15^2) - rac{1}{2}(60)v_i^2$ $-1250 = 6750 - 30v_i^2$ $30v_i^2 = 6750 + 1250$ $v_i^2 = 250$ $v_i = ext{√}(250) ext{ m/s} ext{ or } 15.81 ext{ m/s}$

Step 4

Calculate the average power dissipated by the skateboarder to overcome friction between P and Q.

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Answer

Average power ( $P_{ ext{avg}}$ ) is calculated using the formula:

$P_{ ext{avg}} = rac{W}{ ext{time}}$

The work done against friction over the distance (25 m) is: $W = 50 ext{ N} imes 25 ext{ m} = 1250 ext{ J}$
The time taken ( $ext{time}$ ) to travel from point P to Q: $ext{time} = rac{d}{ ext{speed}} = rac{25 ext{ m}}{15 ext{ m/s}} = 1.67 ext{ s}$
Now calculating average power: $P_{ ext{avg}} = rac{1250}{1.67} ext{ W} ext{ or } 748.5 ext{ W}$

5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q - NSC Physical Sciences - Question 5 - 2017 - Paper 1

Worked Solution & Example Answer:5.1 Draw a labelled free-body diagram, showing ALL the forces acting on the boy-skateboard unit while moving down the ramp from P to Q - NSC Physical Sciences - Question 5 - 2017 - Paper 1

Draw a labelled free-body diagram

State the work-energy theorem in words.

Use energy principles to calculate the speed $v_i$ of the skateboarder at point P.

Calculate the average power dissipated by the skateboarder to overcome friction between P and Q.

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