A spring of negligible mass stretches 3.00 cm from its relaxed length when a for

Question

A spring of negligible mass stretches 3.00 cm from its relaxed length when a force of 7.50 N is applied. A 0.500-kg particle rests on a frictionless horizontal surface and is attached to the free end of the spring. The particle is pulled horizontally so that it stretches the spring 5.00 cm and is then released from rest at t = 0. If the spring were replaced with a spring with a spring constant of 125 N/m while keeping everything else the same, what is the new maximum velocity? a) .79 m/s b) 5 m/s c) .31 m/s d) 1.12 m/s

Explanation / Answer

here,

m = 0.5 kg

the elongation , x = 0.05 m

spring constant , k = 125 n/m

let the maximum velocity be v

using conservation of energy

0.5 * m * v^2 = 0.5 * k * x^2

0.5 * v^2 = 125 * 0.05^2

v = 0.79 m/s

the maximum velocity is a) 0.79 m/s

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