Pls help! A block of mass 5 kg is released from rest at a point where the spring (of force constant k= 300 N/m) has its relaxed length. What is the speed of the block after it has fallen a distance of 9 cm?
Given:
The mass of the block is, m=5 kg.
The force constant of spring is k=300N/m.
The falling distance is, y=0.09 m.
When the block falls, the falling distance is equal to the elongation length of the spring.
Thus, the initial mechanical energy of the block is equal to the sum of the spring potential energy and
the final kinetic energy of the block (conservation of energy).
mgy = 0.5 ky? + 0.5 my?
O 6.98 m/s
2.53 m/s
O 15.53 m/s
O 3.00 m/s