Work-Kinetic Energy Theorem question

[tajmann]

A block of mass 12.0 kg slides from rest down a frictionless 35 degress incline and is stopped by a strong spring with a force constant of 3x10^4 N/m. The block slides 3 meters from the point of release to the point where it comes to rest against the spring. When the block comes to rest, how far has the spring compressed?



Net Work= F. change in position - .5v^2


Hello guys. I have been trying to solve this problem but just cannot seem to do it correctly. I set force as mg and the distance traveled as 3 and used distance instead of velocity (as it is shown in an example in the textbook) and solved for d. It doesn't work. Any help would be greatly appreciated.

Thanks

 

[rock.freak667]

The total work done,W=Force in the direction of motion*distance in the direction of the force.
[The force in the direction of motion is not simply mg]


W=The change in kinetic energy+Work done to compress the spring

 

[ephedyn]

Well, the force of gravity acts in the vertical direction. Work done by gravity on the block is the product of the distance traveled and the component of the force in the direction of the force (vertical direction only). You can also look as this as the product of the distance traveled in the vertical direction with the force of gravity.

If you understand this, then you should apprehend that you've got to resolve the vertical component of the distance...

SPOILER:
loss in grav. PE = gain in KE = gain in PE of spring
change in HEIGHT = 3*sin(35), in degrees
loss in PE = mg*3*sin(35)
= 12 * 9.81 * 3 * sin(35)
= PE of spring

Let x denote the linear dimension of compression,
PE of spring = (x^2) * 0.5*3*10^4 = 12*9.81*3*sin(35)

The rest is calculation...

 

[Gear300]

Work associates with the force involved in the direction of motion (meaning the component of the force that allows for the motion). So the force you need to use is not mg altogether, but the component allowing the block to slide down the incline.