2. Consider the system shown below. The masses are: m1= 3.00[kg], m2= 2.00[kg] a

Question

2. Consider the system shown below. The masses are: m1= 3.00[kg], m2= 2.00[kg] and, M= 1.50[kg]. The wheel has a radius R of 0.50[m]. The system starts from rest, the string is very light and doesn’t stretch or slip on the wheel. Calculate: a. The potential energy change of the masses m1 and m2, if the heavier mass lowers its position by 2.00[m]. b. If there is no friction, the total mechanical energy of the system is conserved. How much is the change in the system’s total kinetic energy? c. Calculate the speed of the rising (and falling) mass, as it reaches the 2.00[m] mark.

Explanation / Answer

A. change in PE = m1gh - m2 gh

change in PE = (3-1.5) * 9.8 * 2

change in PE = 29.4 J

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as therse is no friction,

and as enegry is conserved,

change in PE = chnage in KE = 29.4 J

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kE = 0.5 mv^2 = 29.4

v^2 = 2 * 29.4/3

V = 4.42 m/s for raising masss

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kE = 0.5 mv^2 = 29.4

v^2 = 2 * 29.4/1.5

V = 6.62 m/s for falling masss

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