physics problem?

Question

A 2.00-kg block is pushed against a spring with negligible mass and force constant k = 400 N/m, compressing it 0.220 m. When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0.

What is the speed of the block as it slides along the horizontal surface after having left the spring? and i got 3.11 m/s

now i need...How far does the block travel up the incline before starting to slide back down?

Answer 1

Use the energy of the spring --> the maximal gravitanional energy
(1/2)Cu2 = mgh

Th

Answer 2

You do have the correct velocity. To get the incline part, equate the kinetic energy of the block = .5*m*v^2 with potential energy on the incline, which is a function of height only, and is m*g*h. The distance along the incline is then h/sin(37º)

Answer 3

This now becomes an energy/geometry problem

Potential energy: E=h*g*m where m=mass, g=gravity (9.81m/s^2), and h=height.

Kinetic energy: E=.5*m*v^2 where m =mass and v equals velocity.
Set each equation equal to each other. 0.5*m*v^2=h*g*m. Do calculations. 0.5*v^2=h*g. Solve for h. h=(0.5*v^2/g).

Now comes geometry: look at this as a triangle. You now know the height (h). Solve for the hypotenuse to know the distance. The equation for this is h/sin(x)=d where h is height, x is the angle (37) and d is the distance it will travel up the incline.