Determine the percent efficiency

[snowberryhaze]

1. Determine the percent efficiency of the system

www.imageshack.us/photo/my-images/818/photokt.jpg2. Ek=$\frac{mv^2}{2}$
Eg=mgh
W=FΔd
W=ΔE
3. I am assuming energy is lost in this system due to friction possibly.
Also assuming gravity is 9.81m/s^2
The mass of the the suspended object will be known.
Since I am able to use a metre stick, I can measure the dimensions of the ramp and find necessary angles.
Since a string holds the system together, both objects will accelerate at the same speed and travel the same distance.

I know that the potential energy of the hanging mass accelerates the whole system. And when that happens both of the objects gain kinetic energy which decreases the potential energy of the hanging mass. Once the mass hits the ground there will be 0 potential energy. When the suspended mass hits the ground, the cart on the ramp doesn't fall over, it still remains on the ramp.

 

[e^(i Pi)+1=0]

http://imageshack.us/photo/my-images/818/photokt.jpg/



The system will still have PE after the weight hits the ground due to the cart. I'm a bit stumped on this one as well. If you were allowed to use a photo-gate timer or something similar, you could measure the system's KE just before the bob hits the ground and then calculate the system's total final energy. The difference between this and the system's initial PE would be the energy lost, but off the top of my head I'm stumped how to do it without a timer.

 

[snowberryhaze]

I was told that I cannot use anything but a metre stick.

I am also stumped, also the mass is not given for the cart.

 

[Infinitum]

This is quite an interesting question. You can tie a known mass(I believe you can use any known mass) at the end of the plane, and measure the limiting frictional force acting on the cart so that it -just- starts moving. Having the force, you can easily calculate the work done by friction, which would be the energy lost. And ofcourse, all you need is a metre stick

Though, on the downside, this frictional force is not exactly equal to the kinetic frictional force, but I think it would serve as a good approximation.

 

[snowberryhaze]

This is quite an interesting question. You can tie a known mass(I believe you can use any known mass) at the end of the plane, and measure the limiting frictional force acting on the cart so that it -just- starts moving. Having the force, you can easily calculate the work done by friction, which would be the energy lost. And ofcourse, all you need is a metre stick

Though, on the downside, this frictional force is not exactly equal to the kinetic frictional force, but I think it would serve as a good approximation.

So in this experiment I will raise or lower the ramp as I need to to keep the two balanced, than adjust it so that the mass pulls it down?

The force at which it just starts to move is the static force Mewstatic. I am not sure how to link forces together with energy.

If there is a static force acting the system, can I use the conservation of energy and just toss it in there?

Potential decreases = Kinetic energy increases + work being done by friction.

m2=mass of cart
m1=mass of suspended mass

mass of suspended mass x gravity x the amount the cart moves up the ramp - mass of cart x gravity x length it moves down to the ground x sin theta

m2gl - m1glsintheta = 1/2 (m1+m2)v^2 + m2gcostheta x length

So initially gravity is all that acts on the system. There is a force of gravity pulling the suspended mass down to the ground and a force of gravity pulling the cart down the ramp in the opposition direction in the same plane as the ramp.

Afterwards there is kinetic energy as the two masses move together. I don't know if this works but work = change in energy. If there is energy loss in friction, than it must be Force of friction = MewxForce of normal which is also equal to the y-component of the Force of gravity.

With that said I can say that efficiency is equal to the energy output / energy input x 100%

(1 - Mewmass2costheta / m1-m2sintheta ) x 100%

 

[Infinitum]

The force at which it just starts to move is the static force Mewstatic. I am not sure how to link forces together with energy.

Force(call it F) is not the coefficient of friction (μ). The force you get is the weight of the suspended mass which is equal to the frictional force plus the force due to weight of the cart in direction of the incline(call its mass m2, you can measure this by simply hanging the cart and some known masses on the pulley without the inclined plane.), since the system is in equilibrium.

$m_1g = F + m_2gsin\theta$

Then you can find out what the coefficient of friction is...

can I use the conservation of energy and just toss it in there?

Potential decreases = Kinetic energy increases + work being done by friction.

Yep, I'd do that. That would give,

$\Delta PE = \Delta KE + Fd$

Here, F is the frictional force you got, and d is the displacement of the cart from its initial position, along the inclined plane.

PS : be cautious about the signs in your equations

 

[snowberryhaze]

The first equation is equating and isolating for for F (force of friction).

I'm getting m1g=F+m2gcostheta instead of sin. I'm not quite sure how we are using sin instead of cosine.

 

[Infinitum]

Take the component along the inclined plane. That's a sin.

 

[snowberryhaze]

Ahh, okay.

Unfortunately during the lab I couldn't use any other masses. Therefore I could not find the mass of the cart.

 

[Infinitum]

Then, I really don't see a way how I could perform the experiment...