1.60 kg frictionless block is attached to an ideal spring with force constant 315 n/m . initially the spring is neither stretched nor compressed, but the block is moving in the negative direction at 13.0 m/s .
find (a) the amplitude of the motion, (b) the block’s maximum acceleration, and (c) the maximum force the spring exerts on the block.

(a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

Explanation:

Given that,

Mass of block =1.60 kg

Force constant = 315 N/m

Speed = 13.0 m/s

(a). We need to calculate the amplitude of the motion

Using conservation of energy

Put the value into the relation

(b). We need to calculate the block’s maximum acceleration

Using formula of acceleration

Put the value into the formula

(c). We need to calculate the maximum force the spring exerts on the block

Using formula of force

Put the value into the formula

Hence, (a). The amplitude of the motion is 0.926 m.

(b). The block’s maximum acceleration is 182.31 m/s².

(c). The maximum force the spring exerts on the block is 291.69 N.

prolly a beacuse the momentuem the car has increases while going down hill