Hooke's Law and Work Energy THeorem

[ehabmozart]

Homework Statement

At a waterpark, sleds with riders are sent along a slippery,
horizontal surface by the release of a large compressed spring. The
spring with force constant and negligible mass
rests on the frictionless horizontal surface. One end is in contact
with a stationary wall. A sled and rider with total mass 70.0 kg are
pushed against the other end, compressing the spring 0.375 m. The
sled is then released with zero initial velocity. What is the sled’s
speed when the spring (a) returns to its uncompressed length and
(b) is still compressed 0.200 m?

Homework Equations

Using Work Energy THeorem

The Attempt at a Solution

For part b) I attemted the equation ... .5 K (x)^2 = 0.5 m (v^2) and i got an answer of 1.5 but the book attempt was different... they used it with the x of A and got the answer of 2.4 m/s ... What my problem?? Kindly Clarify... THANKS A LOT!

 

[Doc Al]

For part b) I attemted the equation ... .5 K (x)^2 = 0.5 m (v^2)

Use conservation of energy. What's the total energy at any point?

(Setting PE = KE only works in certain cases--such as when all the PE is converted to KE.)

 

[ehabmozart]

But i applied it in the first case, and i got the correct answer?.. I mean what's wrong with the approach as long as there isn't any friction or resistance??

 

[Doc Al]

But i applied it in the first case, and i got the correct answer?.. I mean what's wrong with the approach as long as there isn't any friction or resistance??

It applied in the first case because all the PE was transformed to KE.

Conservation of energy will apply in all cases.

Starting point: Spring fully compressed. What's the spring PE? What's the KE? What's the total energy?

Ending point: Spring compressed by some amount. What's the spring PE? What's the KE? What's the total energy?

 

[ehabmozart]

Allright, can u clarify it using work energy thorem without reffering to potential energy...

 

[Doc Al]

Allright, can u clarify it using work energy thorem without reffering to potential energy...

Work done by the spring = ΔKE of the mass.

 

[ehabmozart]

Great... So work done by force spring to move to the 0.2 is integral of fx dx from 0 to 0.2 which must equalise the difference in kinetic energy where k initial is zero?

 

[Doc Al]

Great... So work done by force spring to move to the 0.2 is integral of fx dx from 0 to 0.2 which must equalise the difference in kinetic energy where k initial is zero?

Almost. The spring moves from x = 0.375 to 0.2 .