Block hitting a spring down an incline

[AnkhUNC]

Homework Statement

In Fig, a block of mass m = 19 kg is released from rest on a frictionless incline of angle θ = 34°. Below the block is a spring that can be compressed 4.5 cm by a force of 210 N. The block momentarily stops when it compresses the spring by 3.6 cm. (a) How far does the block move down the incline from its rest position to this stopping point? (b) What is the speed of the block just as it touches the spring?

http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c08/fig08_41.gif

Homework Equations

The Attempt at a Solution

I found (a) = 0290429321 meters which is correct.
For (b) I tried the following v = sqrt(K*.036m^2)/19kg. But this is also incorrect. Any ideas?

 

[Shooting Star]

I found (a) = 0290429321 meters which is correct.
For (b) I tried the following v = sqrt(K*.036m^2)/19kg. But this is also incorrect. Any ideas?

No decimal point for (a)?

For (b), your approach is correct, but I'm unable to make out if the sqrt covers the 19 in the denominator.

 

[Moetasim]

I would approach this problem by first analyzing the given information and identifying the relevant equations and principles that can be applied. From the given information, we can see that the block is released from rest on a frictionless incline and that the spring below it can be compressed by a force of 210 N. We also know the mass of the block and the angle of the incline. This information can be used to calculate the potential energy and kinetic energy of the block at different points during its motion.

To solve for (a), we can use the conservation of energy principle, which states that the total energy of a system remains constant. At the starting position, the block only has potential energy, which can be calculated using the equation PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the block on the incline. At the stopping point, the block has both potential and kinetic energy, which can be calculated using the equation KE = 1/2mv^2. By equating the potential energy at the starting point to the sum of potential and kinetic energy at the stopping point, we can solve for the distance traveled by the block.

To solve for (b), we can use the equation for the potential energy of a spring, PE = 1/2kx^2, where k is the spring constant and x is the displacement of the spring. We know that the spring is compressed by 3.6 cm, so we can solve for the spring constant using the given force of 210 N. Once we have the spring constant, we can use the equation for kinetic energy to solve for the velocity of the block just as it touches the spring.

In summary, to solve this problem as a scientist, we would use the principles of conservation of energy and the equations for potential energy and kinetic energy to solve for the distance traveled by the block and its velocity just as it touches the spring.