A 7.0-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 1.2 kN/m).The block is initially at rest at its equilibrium position when a force of magnitude P acting parallel to the surface is applied to the block,as shown.When the block is 8.0 cm from the equilibrium position,it has a speed of 0.80 m/s.How much work is done on the block by the force P as the block moves the 8.0 cm?
A)7.4 J
B)5.4 J
C)6.1 J
D)6.7 J
E)4.9 J

Question

A 7.0-kg block on a horizontal frictionless surface is attached to a light spring (force constant = 1.2 kN/m).The block is initially at rest at its equilibrium position when a force of magnitude P acting parallel to the surface is applied to the block,as shown.When the block is 8.0 cm from the equilibrium position,it has a speed of 0.80 m/s.How much work is done on the block by the force P as the block moves the 8.0 cm? 
A)7.4 J 
B)5.4 J 
C)6.1 J 
D)6.7 J 
E)4.9 J

Quizplus · Accepted Answer

From the work-energy principle, the work done by the force P is equal to the change in the block's kinetic energy.

Initially, the block is at rest and its kinetic energy is zero. 
As the block moves 8.0 cm from equilibrium, it gains potential energy stored in the spring:

U = (1/2)kx^2 = (1/2)(1200 N/m)(0.08 m)^2 = 3.84 J

At this point, the block also has kinetic energy:

K = (1/2)mv^2 = (1/2)(7.0 kg)(0.80 m/s)^2 = 2.24 J

The total mechanical energy of the system (spring-block) is conserved, so the work done by the force P is equal to the total energy change:

W = ΔKE = Kf - Ki = 2.24 J - 0 J = 2.24 J

Therefore, the correct answer is:

Choice C) 6.1 J

A 70-Kg Block on a Horizontal Frictionless Surface Is Attached to to a Light