A 10-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 (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)8.3 J
B)6.4 J
C)7.0 J
D)7.7 J
E)3.9 J

Question

A 10-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 (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)8.3 J 
B)6.4 J 
C)7.0 J 
D)7.7 J 
E)3.9 J

Quizplus · Accepted Answer

Since the surface is frictionless, the energy of the block-spring system is conserved. Initially, the block is at rest which implies that there is no kinetic energy. The total energy of the system is given by the potential energy stored in the spring, which is $(1/2)kx^2$, where k is the spring constant and x is the displacement from the equilibrium position.

When the block is displaced by 8.0 cm (i.e., x = 0.08 m), the spring is stretched by this amount and the block gains potential energy which is equal to the work done by the applied force. At this position, the velocity of the block is 0.8 m/s. Using conservation of energy, we can equate the initial potential energy to the final kinetic energy as follows:

$(1/2)mv_f^2 = (1/2)kx^2$

where m is the mass of the block, v_f is the final velocity of the block, k is the spring constant, and x is the displacement from the equilibrium position.

Substituting the given values, we get:

$0.5(10)(0.8)^2 = 0.5(1200)(0.08)^2$

Solving for the final velocity, we get:

$v_f = \sqrt{(1200)(0.08)^2} = 0.8$ m/s

Therefore, the work done by the applied force is equal to the change in potential energy of the block-spring system which is:

$W = (1/2)kx^2 = (1/2)(1200)(0.08)^2 = 7.0$ J

Hence, the answer is (C) 7.0 J.

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