A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m).The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N)acting parallel to the surface is applied to the block,as shown.What is the speed of the block when it is 13 cm from its equilibrium position?
A)0.85 m/s
B)0.89 m/s
C)0.77 m/s
D)0.64 m/s
E)0.52 m/s

Question

A 10-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 0.80 kN/m).The block is initially at rest at its equilibrium position when a force (magnitude P = 80 N)acting parallel to the surface is applied to the block,as shown.What is the speed of the block when it is 13 cm from its equilibrium position? 
A)0.85 m/s 
B)0.89 m/s 
C)0.77 m/s 
D)0.64 m/s 
E)0.52 m/s

Quizplus · Accepted Answer

The force applied to the block is causing it to undergo simple harmonic motion. The amplitude of the motion is the distance from equilibrium position to the block's maximum displacement. In this case, the amplitude is 13 cm or 0.13 m. 
Using conservation of energy, we can find the maximum speed of the block:
Potential energy at equilibrium position = 0
Maximum potential energy = 0.5kx^2 = 0.5(800 N/m)(0.13 m)^2 = 5.32 J
Maximum kinetic energy = potential energy at maximum displacement = 5.32 J
0.5mv^2 = 5.32 J
v^2 = (2(5.32 J))/(10 kg)
v = 0.85 m/s 
Therefore, the answer is A) 0.85 m/s.

A 10-Kg Block on a Horizontal Frictionless Surface Is Attached