Question

The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 5.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm? The spring constant k = 2000 N/m.

Answer 1

The block moving at instant the spring is compressed 15cm is  √[(25m - 45)/m]

What is conservation of energy principle?

When there is no external force acting on the system, the total mechanical energy remains conserved.

K.E i = K.E f + SPE

1/2 mu² =1/2 mv² + 1/2 kx²

Substitute u = 5m/s, k =2000N.m and x =0.15m, we get the final velocity

1/2 mu² =1/2 mv² + 1/2 kx²

v =√[(25m - 45)/m]

Thus, the block moving at the instant the spring has been compressed 15 cm is √[(25m - 45)/m]

Learn more about conservation of energy principle.

https://brainly.com/question/13097206

#SPJ1