Question

A 3.00 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 N/m. The block is shoved parallel to the spring axis and is given an initial speed of 11.9 m/s, while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion

Answer 1

Answer:

$A=0.185m$

Explanation:

From the question we are told that

Mass of Block $M=3*10^-2kg$

Spring constant $\mu=124 N/m$

Initial Speed $V_1=11.9m/s$

Generally the equation for Kinetic Energy is mathematically given by

 $K.E=1/2 mv^2$

Therefore

 $K.E_1=1/2*0.03*11.9^2$

 $K.E_1=2.12J$

Where

 $Initial K.E_1=Final K.E$

Generally the equation for Final Energy stored is mathematically given by

 $K.E_2=0.5\mu a^2$

Therefore

 $K.E_1=K.E_2$

 $0.5\mu A^2=2.12J$

 $A=\sqrt{\frac{2.12}{0.5*124}}$

 $A=0.185m$

Therefore the amplitude of the resulting simple harmonic motion is

 $A=0.185m$