Solving Pulley System Dynamics: Acceleration & Tension

[chukie]

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*9.80-0.2*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!

 

[anathema]

Hello! It looks like you have the right idea for finding the acceleration in part A. Your calculation for the tension of the rope is also correct. However, for part B, you should use the total mass of the system (12.5 kg) in your calculation for the acceleration. Also, remember that the friction force will act in the opposite direction of the motion, so you should subtract it from the net force. Your final answer for the tension of the rope is correct.

For part C, the downward force of 25 N will act on the remaining block and change the net force acting on it. Your calculation for the acceleration is correct, but remember that the force of friction will also be present in this scenario and will affect the acceleration of the block. So, you will need to consider the friction force in your calculation for the net force. I hope this helps!

 

[Moetasim]

Your calculations for the acceleration and tension in part A are correct. For part B, your calculation for the acceleration is also correct. However, for the tension, you need to use the new mass of 10 kg instead of 2.5 kg. So the correct calculation would be:

T = (10*0.392) + (0.2*10*9.8)/10
T = 4.92 + 1.96
T = 6.88 N

For part C, your calculation for the acceleration is correct. However, since the hanging mass is replaced with a downward force of 25 N, the tension in the rope will also change. The correct calculation for the tension would be:

T = 25 + (0.2*10*9.8)/10
T = 25 + 1.96
T = 26.96 N

I hope this helps clarify your calculations. Keep up the good work!