Question

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2. A catapult throws a stone of mass 0.10 kg with a velocity of 30 m/s2. If 25% of the PE. of the elastic band is wasted during transmission, find the magnitude of the potential energy.

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Answer 1

The mass of the stone (m) = 0.1kg

Velocity of the stone (v) = 30m/s

As you know, kinetic energy of the stone (KE) = 0.5mv2 =0.5 × 0.1× (30)2 = 45J

The kinetic energy of the stone = 25% less than the potential energy of the catapult.

That is, KE (stone) = 75% of PE (catapult)

45J = 0.75× PE

Therefore, PE = 45/0.75 =60J

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Answer 2

Given:

Mass of the stone (m) = $\sf{0.10\:kg}$

Velocity of the stone (v) = $\sf{30\:m/s^{2}}$

As we know:

K. E. of the stone (KE):

$\implies$ $\sf{0.5\:mv^{2}}$

$\implies$ $\sf{0.5 \times 0.10\times (30)^{2}}$

$\implies$ $\sf{45\:J}$

Note: Kinetic energy of the stone is 25% less than the potential energy of the catapult.

That is:

$\boxed{\sf{KE \:(stone) = 75 \%\:of \:PE\:(catapult)}}$

Now:

$\boxed{\sf{45\:J = 0.75 \times PE}}$

Therefore:

Potential energy:

$\implies$ $\sf{ \frac{45}{0.75}}$

$\implies$ $\sf{60\:J}$

Final answer: 60 J