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Three masses m1= 5.0 kg ; m2= 10.0 kg ; m1= 15.0 kg are attached by a strings over frictionless pulleys. The horizontal surface exerts a force of friction of 30N on m2. If the system is released from rest, use energy concepts to find the speed of m2 after it moves 4.0 meter
You need to describe the figure that goes with this. Are m1 and m3 hanging from pulleys at the ends of the horizontal surface? Is 15kg supposed to be m3? You call it m1.
The energy concepts you can use are to equate
Potential energy decrease = (frictional work done) + (kinetic energy increase).
The frictional work done is 30Nx4m = 120 J.
Get the new velocity V from the kinetic energy increase. I assume all three masses will have that velocity.
If one mass rises while another falls an amount X, the potential energy decrease will be (mass difference)* g * X
Sorry there was a mistake on my part
M1=5 kg
M2=10 kg
M3=15 kg
while m1 and m3 are hanging with a support of a pulley and M2 reside on the horizontal surface
Three masses m1= 5.0 kg ; m2= 10.0 kg ; m1= 15.0 kg are attached by a strings over frictionless pulleys. The horizontal surface exerts a force of friction of 30N on m2. If the system is released from rest, use energy concepts to find the speed of m2 after it moves 4.0 meter
You need to describe the figure that goes with this. Are m1 and m3 hanging from pulleys at the ends of the horizontal surface? Is 15kg supposed to be m3? You call it m1.
The energy concepts you can use are to equate
Potential energy decrease = (frictional work done) + (kinetic energy increase).
The frictional work done is 30Nx4m = 120 J.
Get the new velocity V from the kinetic energy increase. I assume all three masses will have that velocity.
If one mass rises while another falls an amount X, the potential energy decrease will be (mass difference)* g * X
Sorry there was a mistake on my part
M1=5 kg
M2=10 kg
M3=15 kg
while m1 and m3 are hanging with a support of a pulley and M2 reside on the horizontal surface
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