Kinetic energy and work

1. Oct 6, 2011

steve357

I have 2 doubts regarding work and KE
first deltaKE=work done which = force * disp * cos theta..if theta lies between 90 and 180 then work done is negative so change in KE is -ve so even final KE can come negative..but we know that KE can never be negative..where am i wrong??Second- how do we find the work done when a variable force changing even its direction at times..acts on a body?

2. Oct 6, 2011

grzz

Work is a scalar just like energy. The + or - sign just shows which part of the system does work on the other part. And the work done may result in KE or KE may be used for some work to be done.

3. Oct 6, 2011

steve357

But how do we find work done when a force changes its direction

4. Oct 6, 2011

HallsofIvy

By integrating along the curved path. If the force is $\vec{F}(x,y,z)$ and the object subjected to the force moves along path (x(t), y(t), z(t)), where "t" is a parameter (and might be time), then the differential along the path is the vector differential, $((dx/dt)\vec{i}+ (dy/dt)\vec{j}+ (dz/dt)\vec{k})dt$. Integrate the dot product of those.

5. Oct 6, 2011

nasu

For the first part, negative work results in negative change in KE. This means that final KE is less than initial KE. Does not mean final KE is negative.

6. Oct 6, 2011

steve357

Yeah that doubt got cleared thanks

7. Oct 6, 2011

steve357

Another doubt- in a question i read that if a block be connected to a spring then it can have constant velocity even in the state of free fall..can it be? and if it so will not the potential energy(spring potential) change throughout the journey?

8. Oct 6, 2011

Staff: Mentor

Where did you read this? Post the full question so we know the context.

9. Oct 7, 2011

steve357

Two blocks A and B are connected to each other by a string and spring (the part from pulley to block b is a spring sry i could not show!).the string passes over a frictionless pulley.Block A slides over the horizontal top surface of a stationary block C and the block B slides along the vertical side of C both with uniform speed.μ is given 0.2 and spring constant 1960Nm-1...B is 2kg calculate mass of A and energy stored in the spring

that's the question since you asked.
However i think my doubt is cleared most probably the spring potential energy shall remain constant whereas the gravitational potential energy shall decrease throughout which shall change into the KE of the Block B conserving the mechanical energy of the system

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10. Oct 7, 2011

Staff: Mentor

Good. Now your question is clear.
Right. The tension is constant (in string and spring) so the spring potential energy will be constant as well.
Careful here. Both blocks are sliding at the same constant speed and there is friction. The kinetic energy of the blocks doesn't change. So as block B falls, what happens to that gravitational potential energy? Is mechanical energy conserved?

11. Oct 8, 2011

steve357

I don't know..nothing about mechanical energy conserved or otherwise is mentioned in the question..But i strongly feel this case is not even possible

See for a moment just consider this case-

you connect a block with a spring,connect the other end of the spring with the ceiling..
now due to gravitational force spring will stretch(until work done by spring force just cancels all the KE of block that increased) and now at this point you disconnect the spring from the ceiling then will the system move with constant velocity? how can it be? because the restoring spring potential energy at that point is more than mg the spring will come back to its natural length and the system will move with acceleration=g. Similarly i think that the case of "spring and block moving with 0 net force" is not possible

Please point out any mistake you can find.

12. Oct 8, 2011

Staff: Mentor

I don't see how this relates to the problem you described in post #9. In that problem, imagine that there is no spring, just a string connecting the two masses. Would you agree that there's no problem in having the masses move at constant speed (if the right mass is chosen)? Would you not further agree that the string must have some tension? (And it should be easy to figure out what that tension must be.) Now if you put back the spring it will have the same tension and thus some degree of stretch and some stored energy. Make sense?

13. Oct 8, 2011

steve357

Fine that's alright but where does the potential energy of the string go then?Since their's no form visible to me i can with much confidence say that your string case is also not possible.

14. Oct 8, 2011

Staff: Mentor

What do you mean by 'potential energy of the string'? It's massless and has no gravitational PE or elastic PE. (Or did you mean to say spring?)

When you include the spring, sure it gets stretched, which requires some energy to do. But we don't care how the spring got stretched, just that it is stretched. And, since the tension remains constant, so will the spring PE.