Question

The 7.3 kg object is released from rest at a
height of 6.2 m on a curved frictionless ramp.
At the foot of the ramp is a spring of force
constant 595 N/m. The object slides down
the ramp and into the spring, compressing it
a distance x before coming momentarily to
rest. The acceleration of gravity is 9.81 m/s
2
Find x

Answer 1

complete the question buddy

Answer 2

I did

Answer 3

Conservation of energy is key here.

Let's realize that the potential energy of the mass before it is dropped will be turned entirely into kinetic energy right before it impacts the spring.

Also realize, that all this kinetic energy will stored in the string. From the work-energy theorem, we can find an expression for the energy stored in a spring. \[W = \int\limits F dx = \int\limits kx dx = {1 \over 2} kx^2 = E_s\]

Do you follow so far?

Answer 4

Therefore, the potential energy before the mass is released can be related to the energy stored in the spring as\[mgh = {1 \over 2} kx^2\]We can readily solve for x.

Answer 5

\[x = \sqrt{\frac{2mgh}{k}}\]