A 263-g block is dropped onto a vertical spring with force constant k = 2.52N/cm. The block sticks to the spring, and the spring compress 11.8 cm before coming momentarily to rest. while the spring is being compressed, how much work is done:

Question

(a) By the force of gravity
(b) By the spring?
(c) What was the speed of the block just before it hit the spring?
(d) If this initial speed of the block is doubled, what is the maximum compression of the spring? Ignore friction.

Answer 1

Ke of block = (1/2) m v^2
when spring is uncompressed

(a) work done by gravity on mass is m g h
= .263*9.81*.118 Joules

(b) work done by spring on mass is -(1/2)kx^2 = -.5(252 N/m)(.118)^2

(c) so because v = 0 at bottom
initial Ke + work done by g = work done on spring by mass
.5 (.263)v^2 + .263*9.81*.118 = .5(252 N/m)(.118)^2
solve for v

now do d :)