Conservation of Energy With Friction

[jakeginobi]

Homework Statement

A 10 kg initially at rest is pulled 13m across a floor by a 50N force. if friction does 380J of work over this distance, what is the block's final velocity? http://imgur.com/a/zM1MX

Homework Equations

W=Fd, Ek=1/2mv^2

The Attempt at a Solution

Since the block was at initially at rest the Energy initially is 0, so I set up the equation like this 0 = Ek after + Efriction. Since friction is 380J i moved it to the other side, then did 1/2mv^2 = 380J to find V which I got 8.72 m/s, but the answer is 7.3m/s. Did I set up the equation wrong?

 

[kuruman]

You forgot to include the 50 N force in your calculation.

 

[jakeginobi]

You forgot to include the 50 N force in your calculation.

How would the entire equation look like in terms of Energy like Ek before = Ek after - E friction?

 

[kuruman]

It is the work-energy theorem.

KE_final - KE_initial = W_friction + W_{50 N force}

You need to find expressions for each one of the terms and put it together.

 

[jakeginobi]

It is the work-energy theorem.

KE_final - KE_initial = W_friction + W_{50 N force}

You need to find expressions for each one of the terms and put it together.

Oh okay thank you. Last question is work and energy the same thing, and are they both scalar quantities?

 

[haruspex]

Oh okay thank you. Last question is work and energy the same thing, and are they both scalar quantities?

Yes, they're both scalars. Energy is the more generic term, while work is used for energy transfer.
In thermodynamics, work refers to energy transfer in mechanical form at the macroscopic level, while heat refers to energy transfer at the level of uncoordinated motion of molecules. Exactly where one draws the line is not defined, and may depend on context.
In regard to this thread, note that in kuruman's equation both the work terms on the right refer to work done on the object. They will not both be positive.

 

[jakeginobi]

Yes, they're both scalars. Energy is the more generic term, while work is used for energy transfer.
In thermodynamics, work refers to energy transfer in mechanical form at the macroscopic level, while heat refers to energy transfer at the level of uncoordinated motion of molecules. Exactly where one draws the line is not defined, and may depend on context.
In regard to this thread, note that in kuruman's equation both the work terms on the right refer to work done on the object. They will not both be positive.

http://imgur.com/a/VHZyU For a question like this, does it matter which direction the velocity is? would I take the resultant velocity to find kinetic energy, so 1/2(m)(500m/s)^2? Or do I need to break it into x and y components?

 

[haruspex]

would I take the resultant velocity to find kinetic energy, so 1/2(m)(500m/s)^2?

I assume you mean for the initial KE. Yes, try that.

 

[jakeginobi]

Yes, they're both scalars. Energy is the more generic term, while work is used for energy transfer.
In thermodynamics, work refers to energy transfer in mechanical form at the macroscopic level, while heat refers to energy transfer at the level of uncoordinated motion of molecules. Exactly where one draws the line is not defined, and may depend on context.
In regard to this thread, note that in kuruman's equation both the work terms on the right refer to work done on the object. They will not both be positive.

Sorry last question, about kuruman's equation, why won't both be positive since they're both scalars I thought direction wouldn't affect it?

 

[jbriggs444]

Sorry last question, about kuruman's equation, why won't both be positive since they're both scalars I thought direction wouldn't affect it?

Scalars can be positive or negative. The direction of the vector quantities that go into calculating a scalar may (or may not) affect the sign of the scalar. So the general answer is "it depends".

For kinetic energy, $E=\frac{1}{2}mv^2$, it turns out that one is taking the scalar product of a vector $\vec v$ with itself and multiplying the result by $\frac{1}{2}m$. Direction does not matter. But in the case at hand, I think you are asking not about kinetic energy, but about work.

We are considering the work done by friction and the work done by the 50N force. Is the work done by friction on the 10 kg object positive or negative? Is the work done by the 50N force on the 10 kg object positive or negative?

 

[kuruman]

We are considering the work done by friction and the work done by the 50N force. Is the work done by friction on the 10 kg object positive or negative? Is the work done by the 50N force on the 10 kg object positive or negative?

And remember, the work done by a constant force on an object is the product of three quantities, the magnitude of the force doing the work, the magnitude of the displacement of the object and the cosine of the angle between the force and the displacement. So, under what circumstances is the work done by the force on the object positive, zero or negative?