A spring with a stiffness constant of 200\text{ N m}^{-1} is compressed by 0.3\text{ m} on… — JAMB UTME

A spring with a stiffness constant of $200\text{ N m}^{-1}$ is compressed by $0.3\text{ m}$ on a frictionless horizontal table. A block of mass $0.5\text{ kg}$ is placed against the compressed spring. When the system is released from rest, all the stored elastic potential energy of the spring is transferred to the block. What is the speed of the block as it leaves the spring?

$6.0\text{ m s}^{-1}$ Answer
B
$36.0\text{ m s}^{-1}$
C
$18.0\text{ m s}^{-1}$
D
$3.0\text{ m s}^{-1}$

Answer

The speed of the block as it leaves the spring is

6.0\text{ m s}^{-1}

By the law of conservation of energy, the elastic potential energy stored in the spring when compressed by

x = 0.3\text{ m}

E_p = \frac{1}{2}kx^2 = \frac{1}{2}(200)(0.3)^2 = 9\text{ J}

. Upon release, this energy converts fully into the block's kinetic energy

E_k = \frac{1}{2}mv^2 = 9\text{ J}

. Substituting

m = 0.5\text{ kg}

gives

0.25 v^2 = 9

, so

v^2 = 36

and

v = 6.0\text{ m s}^{-1}

Step-by-Step Solution

Calculate the elastic potential energy (

E_p

) stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 200 \times (0.3)^2 = 100 \times 0.09 = 9.0\text{ J}

Energy stored in a compressed ideal spring is given by Hooke's law energy formula.

Apply the law of conservation of mechanical energy to find the kinetic energy (

E_k

) of the block.

E_k = E_p = 9.0\text{ J}

On a frictionless surface, all elastic potential energy converts entirely into translational kinetic energy.

Solve for the velocity (

v

) using the kinetic energy formula

E_k = \frac{1}{2} m v^2

9.0 = \frac{1}{2} (0.5) v^2 \implies 0.25 v^2 = 9.0 \implies v^2 = 36 \implies v = 6.0\text{ m s}^{-1}

Isolating

v

requires dividing by half the mass and taking the principal square root.

Key Concept

Conservation of Mechanical Energy (Elastic Potential Energy to Kinetic Energy)

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