Question

when a mass of 2kg is suspended slowly by a spring with its upper end fixed to the ceiling,the spring stretches by 1cm.The work done by the spring is(g=10ms^-2).

Answer 1

Answer:

20J will be the ans....

Answer 2

Answer:

The work done by the spring is 0.1 Joule.

Explanation:

The work done by the spring is calculated as,

$w=\frac{1}{2}kx^2$     (1)

Where,

w=work done by the spring

k=spring constant

x=extension in the spring

From the question we have,

A mass suspended from the spring=2kg

The extension in the spring=1cm=0.01m

The acceleration due to gravity(g)=10m/s²

We know,

$F=kx$        (2)

$F=mg$      (3)

F=force acting on the spring

By using equation (2) in equation (1) we get;

$w=\frac{1}{2}Fx$     (4)

Now by using equation (3) in equation (4) we get;

$w=\frac{1}{2}mgx$     (5)

By inserting the values in equation (5) we get;

$w=\frac{1}{2}\times 2\times 10\times 0.01$

$w=0.1\ Joule$

Hence, the work done by the spring is 0.1 Joule.

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