Question

A toy of mass 0.155 kg is undergoing simple harmonic motion (SHM) on the end of a horizontal spring with force constant 305 N/m . When the object is a distance 1.25×10−2 m from its equilibrium position, it is observed to have a speed of 0.305 m/s . Part A What is the total energy of the object at any point of its motion

Answer 1

`310.38* 10^(-4)j`

Step-by-step explanation:

We have given mass m=0.155 kg

Force constant K = 305 N/m

Distance
`X=1.25* 10^(-2)m`

Velocity
`v=.305 m/sec`

The total energy at any position of the motion is give by
`E=(1)/(2)mv^2+KX^2` here
`(1)/(2)mv^2` is energy due to motion and
`KX^2` is energy due to spring elongation

So total energy
`E=(1)/(2)* 0.155* 0.305^2+305* 0.0125^2=310.38* 10^(-4)j`