Q#76: [Conservation of energy > Potential and Kinetic] (Past Exam Paper – June 2014 Paper 12 Q15)
Small mass is placed at point P on inside surface of smooth hemisphere. It is then released from rest. When it reaches lowest point T, its speed is 4.0 m s–1.
Diagram shows speed of mass at other points Q, R and S as it slides down. Air resistance is negligible.
Mass loses potential energy E in falling from P to T.
At which point has the mass lost potential energy E / 4?
A Q B R C S D none of these
Solution 76:
Answer: B.
This question is more easily seen in terms of kinetic energy.
From conservation of energy,
Potential energy (PE) lost in falling from P to T= Kinetic energy (KE) of mass at T
E = ½ m(4)2 = 8m.
Since the mass is at rest at P, its KE is zero at P (its energy is entirely PE at P) and since T is the lowest point, PE at T is zero. So, the total energy of the mass in the system is E {this is also the total PE at P}.
In general, PE lost at a point = KE at that point (since mass is at rest at P)
(Note that even if the mass has lost some PE at any specific point, it may still have some PE at that point)
From above, E = 8m. So, a PE lost of E /4 = 2m.
At Q, KE = ½ m(1)2 = 0.5m = PE lost.
At R, KE = ½ m(2)2 = 2m = PE lost.
Therefore, when the mass is at R, it has half the final speed and so has gained a kinetic energy of E/4.
This is when it has lost a potential energy of E/4.
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