Kinetic Energy on Pulley/Incline System

[spacecataz]

Hey, this is my first time posting on PhysicsForums! I usually go through the archives but I couldn't find a problem like this.

Two blocks m1 and m2 with masses 50 kg and 100 kg respectively are connected by a string over a pulley that is frictionless with negligible mass. The 50 kg block slides on a 37 degree incline that has a coefficient of kinetic friction of .25. The system is released from rest with a force of 25 N pulling down on the 100 kg block. Calculate the change in kinetic energy of block m1 as it moves a distance of 20 m up the incline.

I think I'm supposed to use this:
\Delta K + \Delta U_{g1}+\Delta U_{g2} = W_{friction} + W_{applied}

So far I have
W_{friction} = -\mu m_{1}gcos(\Theta)d
W_{applied} = Fd
\Delta U_{g1} = m_{1}gdsin(\Theta)
\Delta U_{g2} = m_{2}gd

The answer is supposed to be 4090 J but when I crank out the numbers I don't get that.
What am I missing?! I've stared at this problem for too long. Thank for any help.

 

[rl.bhat]

When the systen is released from the rest, they must have the same acceleration. Using free body diagram find the acceleration.
You have given the answer.But you didn't show any calculation. How should I where you are missing?

 

[spacecataz]

Here are the numbers I get...
Wf = -1956.6
Ug1 = 5897.8
Ug2 = 19600
Wapp = 500

Also, doing what you said, I found the acceleration to be 5.88 which I used kinematics to find the velocity and then the kinetic energy but got 1150.

What's the problem!?

 

[catkin]

Some information is missing; where is m2? Dangling off the pulley?

 

[Doc Al]

I think I'm supposed to use this:
\Delta K + \Delta U_{g1}+\Delta U_{g2} = W_{friction} + W_{applied}

That should work, as long as you realize that \Delta K is the KE of both masses.

So far I have
W_{friction} = -\mu m_{1}gcos(\Theta)d
W_{applied} = Fd

Good.

\Delta U_{g1} = m_{1}gdsin(\Theta)
\Delta U_{g2} = m_{2}gd

That second one should be negative, since m2 moves down.

 

[rl.bhat]

Wfriction = mu*m1*g*sin(theta)*d. Component of g along the inclined plane is
gsin(theta). And the pulling force must be 25N in addition of its weight.Try again.

 

[Doc Al]

Wfriction = mu*m1*g*sin(theta)*d. Component of g along the inclined plane is
gsin(theta).

No. The friction force is mu*N = mu*m1*g*cos(theta).

And the pulling force must be 25N in addition of its weight.

No. By considering gravitational PE, the effect of the weight is automatically included.

 

[rl.bhat]

Yes.You are right.

 

[rl.bhat]

Also, doing what you said, I found the acceleration to be 5.88 which I used kinematics to find the velocity and then the kinetic energy but got 1150.
If you assume that 25N force is applied to 100kg mass by adding some mass to it. And that mass will be 25/9.8 = 2.55kg. So by taking total mass as 102.55kg I got the acceleration as 5.79ms^-2. 50 kg mass start from rest and covers 20m with acceleration 5.79ms^-2. Using the formula
v^2 = u^2 + 2as = 2*5.79*20. And KE = 1/2*50*2*5.79*20. = 5790J.

 

[Doc Al]

Also, doing what you said, I found the acceleration to be 5.88

Please show how you got that acceleration. (I get a different value.)

If you assume that 25N force is applied to 100kg mass by adding some mass to it. And that mass will be 25/9.8 = 2.55kg.

An applied force of 25N is not equivalent to adding 2.55 kg of mass. This is an incorrect approach.