Question

A block of mass 200 g is attached at the end of a massless spring of spring constant 50 N/m. The other end of the spring is attached to the ceiling and the mass is released at a height considered to be where the gravitational potential energy is zero. (a) What is the net potential energy of the block at the instant the block is at the lowest point? (b) What is the net potential energy of the block at the midpoint of its descent? (c) What is the speed of the block at the midpoint of its descent?

Answer 1

Answer:

Explanation:

At the lowest point velocity is zero

loss of potential energy = gain of spring energy

= mgh = 1/2 k h² , h is vertical downward displacement , k is spring constant

2 mg = k h

h = 2mg / k

= (2 x .2 x 9.8) / 50

= .0784 m

P E ( gravitational) = - mgh

= - .2 x 9.8 x .0784

= -  .1536 J

spring PE = + .1536 J

Total PE = 0

b )

At mid point ie at h = .0392 m

gravitational PE = .2 X 9.8 X .0392

= - .0768 J

Elastic PE = 1/2 X 50 X .0392² = .0384 J

Total =  - .0768 J + .0384 J

= - .0384 J

At mid point total energy = 0

- .0384 + KE = 0

KE = .0384 J

c )

1/2 m v ² = .0384

v² = 2 x .0384 / .2

= .384

v = .6196 m / s

62 cm / s