Pulley friction dissipation

[rasen58]

As shown in the figure, a 2kg mass and a 3kg mass hanging from the massless pulley are released from rest. After the 3 kg mass has fallen 1.5 m, it is moving with a speed of 3.8 m/s. What is the rate at which the frictional force dissipates energy during this time interval?

 

My solution:

The 3 kg mass has Fg down and a Tension force up and the 2kg mass has a tension force to the right and a friction force to the left. With my coordinate system, the tension forces cancelled out, and I got

Fnet = Fg - Ffr

Ffr = Fg - Fnet

 

Fg = 3 * 10 = 30 N

Fnet = Mtotal * a

To find the acceleration of the system, I knew V_i = 0m/s, V_f = 3.8 m/s, y = 1.5m

Vf² = V_i²+ 2ay

Then by solving for a, I got 4.813 m/s^2

So now I know Fnet = (5 kg)*4.813 = 24.065 N

And Ffr = 30 N - 24.065 N = about 6 N

 

To find the dissipation of energy, power = work/time

work = force*displacement

The force here is the friction force = 6 N

And I have to find displacement and time now.

 

So to find time, looking at the 3 kg block

y = V_it + 1/2 at²

Then solving for t gave me .7895 seconds.

 

So in .7895 seconds, the 3kg block fell 1.5 meters, so now I need to find how much the 2kg block moved in that time period.

So x = V_it + 1/2 at²

I use v initial = 0 m/s again (right?)

and the same a that I found along with the t = .7895

And got 1.5 meters.

 

So now to find the power dissipated, (6 N * 1.5 meters)/.7895 s = 11.39 Watts

 

But the actual answer is 10 Watts. What did I do wrong?

Thanks.

 

[studiot]

Yes I can confirm that the power is 10.19 watts, by another method.

 

PE lost by 3kg mass = KE gained by system + Work done against friction.

 

From the figures given

 

1.5*3*9.81 = 0.5*5*(3.8)² + W_F

8.045J

 

Power = W_F/t

 

P = 8.045 *3.8/3 = 10.19watts.

 

I agree with your acceleration and time calculations, again by a slightly different method.

 

I am looking more closely at your method to see if I can spot the error.

Edited January 15, 2015 by studiot

[imatfaal]

...

 

I am looking more closely at your method to see if I can spot the error.

 

 

g = 9.81m/s^2 v g =10m/s^

Plus rounding force off too early