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Q.19

Expert-verifiedFound in: Page 256

Book edition
4th

Author(s)
Randall D. Knight

Pages
1240 pages

ISBN
9780133942651

A 10 kg runaway grocery cart runs into a spring with spring constant 250 N/m and compresses it by 60 cm. What was the speed of the cart just before it hit the spring?

As a result, the cart's speed when it hits the spring is $3.0\mathrm{m}\xb7{\mathrm{s}}^{-1}$.

The ratio of an object's distance travelled (in any direction) to the time necessary to travel that distance.

The kinetic energy of a moving object is expressed as,

$T=\frac{1}{2}m{v}^{2}$

we have $T$ is kinetic energy, $m$ is mass, $v$ is velocity.

The phrase for spring's potential energy is as follows:

$U=\frac{1}{2}k{x}^{2}$

we put $k$= spring constant, and$x$ = compression in the spring.

The kinetic energy of the cart before it collides with the spring must equal the potential energy stored in the spring, according to the rule of conservation of energy.

$\frac{1}{2}m{v}^{2}=\frac{1}{2}k{x}^{2}$ we have, $m$= mass , $v$ = velocity, $k$ = spring constant, and $x$= compression in the spring.

Arranging them again for$v$ as follows,

$m{v}^{2}=k{x}^{2}$

${v}^{2}=\frac{k}{m}{x}^{2}$

$v=\sqrt{\frac{k}{m}}x$

The cart's velocity when it hits the spring is computed as follows: Substitute $250\mathrm{N}\xb7{\mathrm{m}}^{-1}$ for $k,10\mathrm{kg}$ for $m$, and $60\mathrm{cm}$ for $x$,

$v=\left(\sqrt{\frac{250\mathrm{N}\xb7{\mathrm{m}}^{-1}}{10\mathrm{kg}}}\right)(60\mathrm{cm})\left(\frac{1.0\mathrm{m}}{100\mathrm{cm}}\right)$

$=3.0\mathrm{m}\xb7{\mathrm{s}}^{-1}$

As a result, the cart's speed when it hits the spring is $3.0\mathrm{m}\xb7{\mathrm{s}}^{-1}$.

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