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(B) The work done $W$ in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. The work done in stretching the spring from an i

Vedclass · Accepted Answer

Less than the work done in stretching it from $20 \ cm$ to $30 \ cm$.

Vedclass · Answer

(B) The work done $W$ in stretching a spring by an extension $x$ is given by $W = \frac{1}{2} k x^2$. The work done in stretching the spring from an initial extension $x_1$ to a final extension $x_2$ is given by $W = \frac{1}{2} k (x_2^2 - x_1^2)$. For the first case ($10 \ cm$ to $20 \ cm$): $W_1 = \frac{1}{2} k (20^2 - 10^2) = \frac{1}{2} k (400 - 100) = 150 k$. For the second case ($20 \ cm$ to $30 \ cm$): $W_2 = \frac{1}{2} k (30^2 - 20^2) = \frac{1}{2} k (900 - 400) = 250 k$. Comparing the two,$W_1 < W_2$. Therefore,the work done in stretching the spring from $10 \ cm$ to $20 \ cm$ is less than the work done in stretching it from $20 \ cm$ to $30 \ cm$.

The work done in stretching a spring from $10 \ cm$ to $20 \ cm$ is:

Similar Questions

The $P.E.$ of a certain spring when stretched from natural length through a distance $0.3\, m$ is $10\, J$. The amount of work in joule that must be done on this spring to stretch it through an additional distance $0.15\, m$ will be ................ $J$.

The potential energy of a long spring when stretched by $2\,cm$ is $U$. If the spring is stretched by $8\,cm$,the potential energy stored in it will be $.......\,U$.

$Assertion$ : Graph between potential energy of a spring versus the extension or compression of the spring is a straight line.
$Reason$ : Potential energy of a stretched or compressed spring is proportional to the square of extension or compression.

$A$ sphere of mass $m$ is attached to a spring of spring constant $k$ and is held in an unstretched position over an inclined plane as shown in the figure. After letting the sphere go,find the maximum length by which the spring extends,given the sphere only rolls.

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