Question

10. Define potential energy. A hammer of mass 1 kg falls on a nail from a height of 2m. How much kinetic energy will it have just before hitting the nail? (g=10m/s2)
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Answer 1

$\Large{\bf \red{What\:is\: Potential\: Energy\:?}}$

The energy which a body has by virtue of its position or configuration in a conservative force field refered to as potential energy .

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$\Large\underline{\underline{\bf \red{Question}:}}$

A hammer of mass 1 kg falls on a nail from a height of 2m. How much kinetic energy will it have just before hitting the nail? (g=10m/s2)

$\Large\underline{\underline{\sf Given:}}$

• Mass of hammer (m) = 1kg
• Height (h) = 2m

$\Large\underline{\underline{\sf To\:Find:}}$

• Kinetic Energy just before hitting the nail.

$\Large\underline{\underline{\sf Formula\:Used:}}$

$\large{\boxed{\boxed{\sf \pink{v^2=u^2+2gs}}}}$

$\large{\boxed{\boxed{\sf \pink{KE=\dfrac{1}{2}mv^2}}}}$

$\Large\underline{\underline{\sf Solution:}}$

$\Large{\sf v^2=u^2+2gs}$

Initial Velocity of hammer is 0m/s

$\implies{\sf v^2=0+2×10×2}$

$\implies{\sf v^2=40 }$

$\implies{\sf v=\sqrt{40}m/s }$

$\Large{\sf Kinetic\: Energy=\dfrac{1}{2}mv^2}$

$\implies{\sf KE=\dfrac{1}{2}×1×\sqrt{40}^2 }$

$\implies{\sf KE=\dfrac{1}{2}×1×40 }$

$\implies{\sf KE=20\:J}$

$\Large\underline{\underline{\sf Answer:}}$

•°• Kinetic Energy just before hitting the nail is 20J .