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byeh2004

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A m1 = 49.0 kg block and a m2 = 100.0 kg block are connected by a string as in Figure P7.28. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 49.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 49.0 kg block as it moves from A to B, a distance of 20.0 m.



I made it into a tension/force problem at first, with the vertical or heavier block being the cause of the tension since the sum of the vertical forces of the tension was 0. So T = mg. So I knew what T was

Then for the block on the incline, I made a free body diagram where mgsin37 and Fk (friction force) pointed towards the left and T (tension) pointed towards the right. For vertical forces on the block, the sum was 0. So I can assume that n=mgcos37.

So my sigma Fx for the block on the incline was
T - mgsin37 + Fk

I calculated it and then plugged it into the work equation.

W= FDcos(theta)

Because the change of kinetic energy is just net work done. I thought W was my final answer.

Can anyone help me? Thanks!
 

DrChandy

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byeh2004 said:
A m1 = 49.0 kg block and a m2 = 100.0 kg block are connected by a string as in Figure P7.28. The pulley is frictionless and of negligible mass. The coefficient of kinetic friction between the 49.0 kg block and incline is 0.250. Determine the change in the kinetic energy of the 49.0 kg block as it moves from A to B, a distance of 20.0 m.



I made it into a tension/force problem at first, with the vertical or heavier block being the cause of the tension since the sum of the vertical forces of the tension was 0. So T = mg. So I knew what T was

Then for the block on the incline, I made a free body diagram where mgsin37 and Fk (friction force) pointed towards the left and T (tension) pointed towards the right. For vertical forces on the block, the sum was 0. So I can assume that n=mgcos37.

So my sigma Fx for the block on the incline was
T - mgsin37 + Fk

I calculated it and then plugged it into the work equation.

W= FDcos(theta)

Because the change of kinetic energy is just net work done. I thought W was my final answer.

Can anyone help me? Thanks!

Forces on M1
Gravity: sinmg=sin37(50kg)(10m/s^2)=300kgm/s^2
+
Kinetic Friction: uN=(0.25)cos37(500kgm/s^2)=100kgm/s^2
Net Force=400kgm/s^2

Force on M2
Gravity: mg=100kg(10m/s^2)=1000kgm/s^2
--------------------------------------------

Net Force on system: 1000kgm/s^2-400kgm/s^2=600kgm/s^2
Net Mass of system: 100kg+50kg=150kg

Therefore net acceleration of system:
F=ma
600kgm/s^2=(150kg)a
a=4m/s^2

--------------------------------------------

Change in kinetic energy of M1:
Vf^2=Vi^2+(Xf-Xi)2a
Vf^2= 0 +( 20m )2(4m/s^2)
Vf^2= 160

1/2(M)Vf^2=
1/2(50kg)160(m/s)^2=
4000kg(m/s)^2


It might help for you to double check the calculations
Good luck
 

byeh2004

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Thanks alot man!

Why do we ignore the other forces like normal force cancelling out gravitational force?
 
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DrBowtie

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byeh2004 said:
Thanks alot man!

Why do we ignore the other forces like normal force cancelling out gravitational force?
If not the block would be accelerating off of the slope.
 

Mr. Tee

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Remember to use 49 kg for m1.

I used this mass and still got the acceleration a= 4.0 m/s^2 like DrChandy did. However, the final answer (change in kinetic energy) will be different from what he stated.
 

DrChandy

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MCAT related concepts. This is a detailed problem from a physics textbook or webassign.
 
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