Answer
a) There are two forces that do nonzero work on the object: the gravitational force and the force exerted by the spring. The gravitational force does work on the object as it is lowered to its equilibrium position because the force is acting in the direction opposite to the displacement of the object. The force exerted by the spring also does work on the object as it is compressed, storing potential energy in the spring.
b) The work done by the gravitational force is given by:
$W_gravity = -mgh$
where m is the mass of the object, g is the acceleration due to gravity, and h is the distance the object is lowered. Since the object is lowered to its equilibrium position, h = 0, and the work done by gravity is zero.
The work done by the spring force is given by:
$W_spring = (1/2)kx^2$
where k is the spring constant and x is the displacement of the object from its equilibrium position. As the object is lowered, x decreases until it reaches zero at the equilibrium position. The work done by the spring force is then:
$W_spring = (1/2)kx_(initial)^2$
where x_initial is the initial displacement of the object before it was lowered to its equilibrium position. Since the object was initially at rest and at its equilibrium position, x_initial = 0, and the work done by the spring force is zero.
Work Step by Step
a) There are two forces that do nonzero work on the object: the gravitational force and the force exerted by the spring. The gravitational force does work on the object as it is lowered to its equilibrium position because the force is acting in the direction opposite to the displacement of the object. The force exerted by the spring also does work on the object as it is compressed, storing potential energy in the spring.
b) The work done by the gravitational force is given by:
$W_gravity = -mgh$
where m is the mass of the object, g is the acceleration due to gravity, and h is the distance the object is lowered. Since the object is lowered to its equilibrium position, h = 0, and the work done by gravity is zero.
The work done by the spring force is given by:
$W_spring = (1/2)kx^2$
where k is the spring constant and x is the displacement of the object from its equilibrium position. As the object is lowered, x decreases until it reaches zero at the equilibrium position. The work done by the spring force is then:
$W_spring = (1/2)kx_initial^2$
where x_initial is the initial displacement of the object before it was lowered to its equilibrium position. Since the object was initially at rest and at its equilibrium position, x_initial = 0, and the work done by the spring force is zero.
