Question

A 1.10 kg block is attached to a spring with spring constant 18.0 N/m . While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 43.0 cm/s . What is the block's speed at the point where x= 0.650 A? (if the amplitude of the subsequent oscillations is 10.6cm

Answer 1

Given : m= 1.10 kg , k= 18 N/m , v= 43 cm/s = 0.43 m/s , A= 10.6 cm ,

x = 0.650 A = 0.650*10.6= 6.89 cm

Solution:

Step:1

Initial kinetic energy is given by :

KE_i = (1/2)mv²

= (1/2)(1.10 kg)(0.43 m/s)²

= 0.1017 J

Step:2

Kinetic energy at point x = 0.650 A = 6.89 cm = 0.0689 m

KE_f = (1/2)mv_f²

= (1/2)(1.10 kg)(v_f)²

= 0.55 v_f²

Potential energy at point x :

PE_f = (1/2)kx²

= (1/2)(18 N/m)(0.0689 m)²

= 0.0427 J

Step:3

Applying conservation of energy :

KE_i = PE_f + KE_f

0.1017 = 0.0427 + 0.55v_f²

By rearranging we get

v_f² = 0.1073

By solving we get , v_f = 0.328 m/s

Answer: Block's speed at point x is 0.328 m/s