Question

A block of mass m = 2.00 kg is attached to a spring of force constant k = 5.35 ✕ 102 N/m that lies on a horizontal frictionless surface as shown in the figure below. The block is pulled to a position xi = 5.15 cm to the right of equilibrium and released from rest.

a) Find the the work required to stretch the spring.
b) Find the speed the block has as it passes through equilibrium.

Answer 1

A; 6.625 J

B; 0.06 m/s

Step-by-step explanation:

The total mechanical energy of the system is conserved, which means the initial potential energy stored in the spring is converted into kinetic energy as the block passes through equilibrium.

a) The work required to stretch the spring is equal to the potential energy stored in the spring at the maximum displacement. The potential energy of a spring can be calculated using the formula:

PE = (1/2) * k * Δx^2

Where PE is the potential energy, k is the force constant, and Δx is the displacement from equilibrium.

Given:

Mass of the block (m) = 2.00 kg

Force constant of the spring (k) = 5.35 × 10^2 N/m

Initial displacement (xi) = 5.15 cm = 0.0515 m

Substituting the values into the formula:

PE = (1/2) * (5.35 × 10^2 N/m) * (0.0515 m)^2

= 6.625 J

Therefore, the work required to stretch the spring is 6.625 Joules.

b) At equilibrium, the block has converted all its potential energy into kinetic energy. The total mechanical energy (E) of the system is given by:

E = (1/2) * m * v^2

Where E is the total mechanical energy, m is the mass of the block, and v is the velocity of the block at equilibrium.

Since the total mechanical energy is conserved, we can equate the initial potential energy to the kinetic energy at equilibrium:

PE = E

(1/2) * k * Δx^2 = (1/2) * m * v^2

Substituting the known values:

(1/2) * (5.35 × 10^2 N/m) * (0.0515 m)^2 = (1/2) * (2.00 kg) * v^2

Simplifying the equation:

(1.3603 N) * (0.00264725 m^2) = v^2

v^2 = 0.00360038665 m^2/s^2

Taking the square root of both sides:

v = 0.06 m/s

Therefore, the speed of the block as it passes through equilibrium is 0.06 m/s.

Answer 2

To stretch the spring, a work of 0.0674 J is required. When the block passes through equilibrium, it has a speed of 0.664 m/s.

To find the work required to stretch the spring, we can use the equation:

Work =
`(1/2)kx^2,`

where k is the force constant and x is the displacement from equilibrium. Plugging in the values, we get:

Work =
`(1/2)(5.35 ✕ 10^2 N/m)(0.0515 m)^2.`

Calculating this, the work required to stretch the spring is 0.0674 J.

To find the speed of the block as it passes through equilibrium, we can use conservation of energy. At the point of release, all of the potential energy is stored in the spring. So the potential energy at equilibrium can be converted to kinetic energy. Using the equation:

`(1/2)mv^2 = (1/2)kx^2,`

where m is the mass of the block, k is the force constant, v is the speed of the block, and x is the displacement from equilibrium, we can solve for v:

`(1/2)(2.00 kg)v^2 = (1/2)(5.35 ✕ 10^2 N/m)(0.0515 m)^2,`

Calculating this, the speed of the block as it passes through equilibrium is 0.664 m/s.