The 0.100-kg sphere (Figure 1) is released from rest at the position shown in th
ID: 2298130 • Letter: T
Question
The 0.100-kg sphere (Figure 1) is released from rest at the position shown in the sketch, with its center 0.400 m from the center of the 5.00-kg mass. Assume that the only forces on the 0.100-kg sphere are the gravitational forces exerted by the other two spheres and that the 5.00-kg and 10.0-kg spheres are held in place at their initial positions.
A) What is the speed of the 0.100-kg sphere when it has moved 0.200 m to the left from its initial position?
Express your answer to three significant figures and include the appropriate units.
Explanation / Answer
The 0.1 kg object can't move to the left !!!
The change in gravitational potential energy is given by:
?PE = G M? M? (1/Ri - 1/Rf)
where
G = gravitational constant = 6.67E-11 Nm/kg
M? = mass of one object
M? = mass of the other object
Ri = initial distance
Rf = final distance
The change in gravitational potential energy by the 10 kg mass is:
?PE10 = (6.67E-11 Nm/kg) (10 kg) (0.1 kg) [1/(0.6 m) - 1/(0.8 m)] = 2.779E-11 J
The change in gravitational potential energy by the 10 kg mass is:
?PE5 = (6.67E-11 Nm/kg) (5 kg) (0.1 kg) [1/(0.4 m) - 1/(0.2 m)] = -8.337E-11 J
The total change in gravitational potential energy is:
?PE = ?PE10 + ?PE5
?PE = (-8.337E-11 J) + (2.779E-11 J)
?PE = -5.558E-11 J
The loss in potential energy is equal to the kinetic energy gained, so
?KE = 5.558E-11 J
So the speed is:
(5.558E-11 J) = (0.1 kg) v / 2
v = 3.33* 10^-5 m/s (answer)
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