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Two masses are attached to a massless, frictionless pulley as shown. (image is attached)
A) Find the acceleration of the two masses and the tension of the rope connecting them.
B) Now the top surface is replaced with one which has a coefficient of kinetic friction of 0.2. FInd the acceleration of the two masses and the tension in the rope.
C) Finally, the hanging mass is replaced with a downward force of 25 N. Find the acceleration of the remaining block.
My Work:
A) a= Fnet/m
a=(2.5*9.80)/(10+2.5)
a=1.96 m/s^2
I am pretty sure my acceleration is right, but I'm not sure about my number for the tension of the rope connecting them:
T=(2.5*9.8)(2.5*1.96)
T=19.6 N
As for B and C, i'm not sure if I did it right.
B) a= Fnet/m
a=(2.5*9.800.2*10*9.8)/(10+2.5)
a=0.392 m/s^2
T=(10*0.392)+(0.2*10*9.8)/10
T= 9.065N
C) a=Fnet/m
a= (25(0.2*10*9.8))/10
a= 0.54
Could someone help me please? Thanks in advance!
A) Find the acceleration of the two masses and the tension of the rope connecting them.
B) Now the top surface is replaced with one which has a coefficient of kinetic friction of 0.2. FInd the acceleration of the two masses and the tension in the rope.
C) Finally, the hanging mass is replaced with a downward force of 25 N. Find the acceleration of the remaining block.
My Work:
A) a= Fnet/m
a=(2.5*9.80)/(10+2.5)
a=1.96 m/s^2
I am pretty sure my acceleration is right, but I'm not sure about my number for the tension of the rope connecting them:
T=(2.5*9.8)(2.5*1.96)
T=19.6 N
As for B and C, i'm not sure if I did it right.
B) a= Fnet/m
a=(2.5*9.800.2*10*9.8)/(10+2.5)
a=0.392 m/s^2
T=(10*0.392)+(0.2*10*9.8)/10
T= 9.065N
C) a=Fnet/m
a= (25(0.2*10*9.8))/10
a= 0.54
Could someone help me please? Thanks in advance!
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