Answer to Question #82614 in Physics for Mia

Question

A 263-g object is dropped onto a vertical spring with force constant k = 2.52 N/cm. The object sticks to the spring, and the spring compresses 11.8 cm before coming momentarily to rest. While the spring is being compressed, how much work is done:
a. by the force of gravity?
b. by the spring?
c. What was the speed of the object just before it hit the spring?
d. If this initial speed of the block is doubled, what is the maximum compression of the spring?

Answer 1

a. The work done by the force of gravity:

W_FG=mgh=0.263·9.8·0.118=0.304 J.

b. The work done by the spring is

W_S=-(kh^2)/2=-(252·〖0.118〗^2)/2=-1.754 J.

c. According to the work-kinetic energy theorem:

ΔK=0-(mv^2)/2=mgh-(kh^2)/2,

v=√(2/m(〖W_FG-W〗_S))=√(2/m (0.304-(-1.754)) )=3.957 m/s.

d. According to the previous theorem:

(kh^2)/2-mgh-(m(2v)^2)/2=0,

126h^2-2.577h-8.236=0,

h=0.266 m.

Answer

a. 0.304 J; b. -1.754 J; c. 3.957 m/s; d. 26.6 cm