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Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 3.21 kg...Asked by Megan
Two blocks are arranged at the ends of a massless string as shown in the figure. The system starts from rest. When the 1.78 kg mass has fallen through 0.438 m its downward speed is 1.27 m/s. The acceleration of gravity is 9.8 m/s^2. What is the frictional force between the 3.13 kg mass and the table? answer in units of newtons.
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Anonymous
change in potential energy = -1.78 (9.81) (0.438) = -7.65 Joules
increase in kinetic energy = (1/2)(1.78+3.13) (1.27)^2 = 3.96 Joules
Oh my, we lost energy = 7.65-3.96 = 3.69 Joules wasted in friction
so
3.69 = F* distance = F * 0.438
F = 8.43 Newtons
increase in kinetic energy = (1/2)(1.78+3.13) (1.27)^2 = 3.96 Joules
Oh my, we lost energy = 7.65-3.96 = 3.69 Joules wasted in friction
so
3.69 = F* distance = F * 0.438
F = 8.43 Newtons
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