A 1.2-kg block sliding at 5.8 m/s on a frictionless surface runs into and sticks to a spring. The spring is compressed 0.10 m before stopping the block and starting its motion back in the opposite direction.

A. Determine the total energy of the block-spring system. Express your answer with the appropriate units.

The total energy of the block-spring system will be 80.74 J.

Explanation:

Given mass (M) of the block is 1.2 Kg. It slides on a frictionless table at a velocity (v) 0f and sticks to a spring and compress the spring by an amount (x) of 0.10m.

So form conservation of energy, we can write

Elastic energy gained by the spring = loss of kinetic energy by the block

If 'k' be the spring constant, then

The total energy (E) of block-spring system will be

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2