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An electric motor pulls a 20 kg crate from rest at point A up an inclined plane by means of a light inextensible rope - NSC Physical Sciences - Question 5 - 2023 - Paper 1
Question 5
An electric motor pulls a 20 kg crate from rest at point A up an inclined plane by means of a light inextensible rope. The inclined plane makes an angle of 18° with ... show full transcript
Worked Solution & Example Answer:An electric motor pulls a 20 kg crate from rest at point A up an inclined plane by means of a light inextensible rope - NSC Physical Sciences - Question 5 - 2023 - Paper 1
Step 1
Define a non-conservative force.
Answer
A non-conservative force is one where the work done on an object depends on the path taken by the object. This means that the energy transferred by the force is not stored as potential energy but is instead transformed into other forms, such as thermal energy. In this case, friction is an example of a non-conservative force, as it does not conserve mechanical energy.
Step 2
Use ENERGY PRINCIPLES to calculate the speed of the crate when it reaches point C.
Answer
To solve for the speed at point C, apply the work-energy principle:
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Calculate total work done:
W_{total} = F_{applied} imes d - F_{friction} imes d
W_{total} = (96.8 ext{ N} - 13.5 ext{ N}) imes 15.6 ext{ m}
W_{total} = 83.3 ext{ N} imes 15.6 ext{ m} = 1300.68 ext{ J} -
The initial kinetic energy (KE_initial) at point A is 0, since it starts from rest.
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Calculate the speed at point C using:
KE_{final} = KE_{initial} + W_{total}
rac{1}{2} m v^2 = 0 + 1300.68
rac{1}{2} (20) v^2 = 1300.68
v^2 = rac{1300.68 imes 2}{20} v^2 = 130.068
v = ext{approximately } 11.4 ext{ m/s}
Thus, the speed of the crate when it reaches point C is approximately 11.4 m/s.
Step 3
Calculate the minimum average power dissipated by the electric motor to pull the crate from point A to point C.
Answer
To find the average power, use the formula:
[ P_{avg} = rac{W_{total}}{t} ]
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We already calculated the work done:
W_{total} = 1300.68 ext{ J} \ -
Calculate time taken (t) to reach point C:
t = \frac{d}{v}
t = \frac{15.6 ext{ m}}{11.4 ext{ m/s}} \approx 1.37 ext{ s} \ -
Now, substitute in for average power:
P_{avg} = \frac{1300.68 ext{ J}}{1.37 ext{ s}} \approx 950.25 ext{ W}
Thus, the minimum average power dissipated by the electric motor is approximately 950.25 W.
Step 4
Draw a velocity-time graph for the entire motion of the crate starting from point A until it passes point B again on its motion down the inclined plane.
Answer
The velocity-time graph would consist of two parts:
- Up the incline (A to C): The graph starts at (0,0) and increases linearly to the top speed of approximately 11.4 m/s at point C.
- Down the incline (C to B): After reaching C, the crate comes to a stop at D, and then accelerates back down to B. The down motion will show a negative slope as it decelerates to a stop at point D and then accelerates down to point B.
(Sketch the graph with velocity on the y-axis and time on the x-axis, clearly indicating the turn at point C and the downward slope to point B.)