A spring with a force constant of 400\text{ N m}^{-1} is compressed by 0.1\text{ m} from i… — JAMB UTME

A spring with a force constant of $400\text{ N m}^{-1}$ is compressed by $0.1\text{ m}$ from its uncompressed length. When released, it launches a block of mass $0.16\text{ kg}$ along a smooth horizontal surface. What is the speed of the block immediately after leaving the spring?

$5.0\text{ m s}^{-1}$ Answer
B
$2.5\text{ m s}^{-1}$
C
$12.5\text{ m s}^{-1}$
D
$25.0\text{ m s}^{-1}$

Answer

The speed of the block immediately after leaving the spring is

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

According to the principle of conservation of mechanical energy, the elastic potential energy stored in the compressed spring (

E_p = \frac{1}{2} k x^2

) is completely converted into kinetic energy (

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

) when the spring is released on a frictionless surface. Substituting

k = 400\text{ N m}^{-1}

x = 0.1\text{ m}

, and

m = 0.16\text{ kg}

yields

2.0\text{ J} = 0.08 v^2

, which solves to

v = 5.0\text{ m s}^{-1}

Step-by-Step Solution

Calculate the elastic potential energy stored in the compressed spring.

E_p = \frac{1}{2} k x^2 = \frac{1}{2} \times 400 \times (0.1)^2 = 200 \times 0.01 = 2.0\text{ J}

Work done in compressing the spring is stored as elastic potential energy.

Apply conservation of mechanical energy.

E_k = E_p \implies \frac{1}{2} m v^2 = 2.0\text{ J}

Since the surface is smooth, all potential energy transforms into kinetic energy.

Solve for the velocity

v

of the block.

\frac{1}{2} \times 0.16 \times v^2 = 2.0 \implies 0.08 v^2 = 2.0 \implies v^2 = 25 \implies v = 5.0\text{ m s}^{-1}

Isolate

v

by dividing by

0.08

and taking the square root.

Key Concept

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

Question

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