Question

A horizontal spring attached to a wall has a force constant of k = 820 N/m. A block of mass m = 1.20 kg is attached to the spring and rests on a frictionless, horizontal surface as in the figure below

(a) The block is pulled to a position x_i = 5.40 cm from equilibrium and released. Find the potential energy stored in the spring when the block is 5.40 cm from equilibrium.
(b) Find the speed of the block as it passes through the equilibrium position.
(c) What is the speed of the block when it is at a position x_i/2 = 2.70 cm?

Answer 1

given
k = 820 N/m
m = 1.20 kg
xi = 5.40 cm = 0.054 m
x = xi/2 = 5.4/2 = 2.7 cm = 0.027 m

a) PE = (1/2)*k*xi^2

= (1/2)*820*0.054^2

= 1.20 J   <<<<<<<<<--------------Answer

b) Apply conservation of energy

kinetic eneegy at equilibrium position = potential energy at maximum extension of the spring.

(1/2)*m*v^2 = (1/2)*k*xi^2

v^2 = k*xi^2/m

v = sqrt(k/m)*xi

= sqrt(820/1.2)*0.054

= 1.41 m/s <<<<<<<<<--------------Answer

c) Again apply conservation of energy

(1/2)*m*v^2 + (1/2)*k*x^2 = (1/2)*k*xi^2

(1/2)*m*v^2 = (1/2)*k*xi^2 - (1/2)*k*x^2

v^2 = (k/m)*(xi^2 -x^2)

v = sqrt((k/m)*(xi^2 - x^2) )

= sqrt( (820/1.2)*(0.054^2 - 0.027^2) )

= 1.22 m/s <<<<<<<<<--------------Answer