Change in Kinetic Energy of block

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Homework Help Overview

The discussion revolves around a physics problem involving the change in kinetic energy of a block moving up an inclined plane. The block's mass, initial speed, distance traveled, and incline angle are provided, prompting participants to explore the relationship between kinetic energy, work done, and potential energy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss various equations for calculating the change in kinetic energy, including Delta K = mgh and Delta K = 1/2mVf^2 - 1/2mVi^2. Questions arise about the correct substitution for height and the interpretation of negative values in energy calculations.

Discussion Status

There is ongoing exploration of different methods to determine the change in kinetic energy and its implications for subsequent calculations involving friction. Some participants have provided clarifications regarding the magnitude of energy changes, while others express confusion about the signs and units in their calculations.

Contextual Notes

Participants note assumptions such as the constancy of the frictional force and the need to differentiate between work done and force in their calculations. There is mention of previous questions regarding potential energy, indicating a broader context of related concepts being discussed.

grouchy
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A 4.5 kg block is set into motion up an inclined plane with an initial speed of 8.9 m/s. The block comes to a rest after traveling 6 m along the plane, which is inclined at an angle of 24 to the horizontal. Determine the magnitude of the change in the block's kinetic energy.

I tried Delta K= Work of the Force
so..

Delta K = mgh but I got it wrong.
 
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Delta K = mgh
What is the value of h you have substituted in this equation?
 
grouchy said:
I tried Delta K= Work of the Force
so..

Delta K = mgh but I got it wrong.
You don't need any such equation to answer the question. Instead: What's the initial KE? What's the final KE? What's the change?
 
I did also try Delta K = 1/2mVf^2 - 1/2 mVi^2. The final is 0 since it is at rest and I got -178.222 for the initial. But that was also wrong
 
rl.bhat said:
Delta K = mgh
What is the value of h you have substituted in this equation?

for h I user 6sin24.
 
grouchy said:
I did also try Delta K = 1/2mVf^2 - 1/2 mVi^2. The final is 0 since it is at rest and I got -178.222 for the initial. But that was also wrong
The initial KE = + 178.22 J. The change in KE is negative. But note that they ask for the magnitude of the change. (So drop the negative sign.)
 
ok, I get that now. Magnitude is just the number and not the sign.

But for the next part it says that (Assumed: The fristional force is constant. Determine the magnitude of the frictional force exerted on the block.)

do I use, Wf = DeltaK + DeltaU?
 
grouchy said:
do I use, Wf = DeltaK + DeltaU?
That will work.
 
I don't get it, I'm getting it wrong. a previous question asked for the change in potential energy which I got to be 107.623. So I think I'm using the wrong sign or something...

I tried -178.223 + 107.623 and got -70.61N. I tried it with and without the negative sign and still got it wrong.
 
  • #10
grouchy said:
I tried -178.223 + 107.623 and got -70.61N. I tried it with and without the negative sign and still got it wrong.
That's the work done by friction, in Joules not Newtons. Now find the friction force.
 
  • #11
man..I suck at this physics stuff lol. I forgot that I found the W and not the force. Thx for your help though!
 

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