Question

A 40.25 pound block on a horizontal frictionless surface is pressed into a spring with a spring constant (k) of 18.0 pounds/inch for a distance of 3.0 inches.

a) What is the potential energy stored in the spring? Answer in ft-lbs.

b) When the block is released, what is the velocity of the block at the moment the block and spring

separate? Answer in ft/sec.

Answer 1

Given : M= 40.25 lbs , K = 18 lbs/in , x = 3 in

Solution:

(a) Potential energy stored in the spring

As the block is pressed in spring , the potential energy stored in the spring is given by:

PE= (1/2)Kx²

= (1/2)(18 lbs/in)(3 in)²

= 81 lbs-in or 6.747 ft-lbs

Answer: PE = 6.747 ft-lbs

(b) Velocity of block when it is released

According to the conservation of energy, when block is released all of its stored potential energy will be tranformed into kinetic energy .

i.e. PE = KE

6.747 ft-lbs = (1/2)(40.25 lbs)(v²)

v² = 0.335

v = 0.335 = 0.579 ft/s

Answer: v= 0.579 ft/s