The complete question is as follows:
A 0.150-kg toy is undergoing SHM on the end of a horizontal spring with force constant k = 300 N/m. When the toy is 0.0120 m from its equilibrium position, it is observed to have a speed of 0.400 m/s.
Answer:
The correct answer is 0.034 J.
Explanation:
Given :
mass of the toy is m = 0.15 kg.
The force constant of restoring force k = 300 Nm⁻¹
When the position of the toy from the equilibrium is x = 0.012m, then the
speed of the toy vx = 0.4m s
The total mechanical energy in SHM is given by
E= 1/2 (mv²+ kx²) = 1/2 kA²
(here, m = mass of the object, vx = velocity, k = force constant
of restoring force, and A = amplitude of SHM.)
Hence by substituting the numerical values in equation 1, we get
[tex]E= \frac{1}{2} (0.15* 0.4) + \frac{1}{2} (300* 0.012)[/tex]
= 0.034 J
Thus, the correct answer is 0.034 J.
