A block of mass 2.95 kg is placed against a horizontal spring of constant k = 820 N/m and pushed so the spring compresses by 0.0800 m. (a) What is the spring potential energy of the block-spring system? 2.624 Correct: Your answer is correct. J (b) If the block is now released and the surface is frictionless, calculate the block's speed after leaving the spring.

(a) 2.624

(b) 1.334

Explanation:

The given data in the question is:

Mass = 2.95 kg

Spring constant "k" = 820 N/m

Compression of Spring "x" = 0.0800m

Using this data and the equation to find the answer for spring potential energy:

spring potential energy = 1/2 kx²

= 1/2 (820 N/m) (0.0800m) ²

= 2.624 Joules

(b) - It is given that the surface is friction-less which means no loss of energy during the motion of spring. This means that all the potential energy possessed by the spring converts to Kinetic Energy of the block. We can equate spring potential energy with Kinetic energy to find the block's speed as below:

Kinetic Energy = 1/2mv² where "m" is mass and "v" is velocity of the object

1/2mv² = spring potential energy (found in part "a" above)

1/2 (2.95 kg) v² = 2.624 Joules

v² = (2.624 x 2) / 2.95

v² = 1.779

v = √1.779

v = 1.334 m/s