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2 years ago

8

Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the bow back 50.

cm and holds it in position with a force of 150 N. If the mass of the arrow is 50. g and the "spring" is massless, what is the speed of the arrow immediately after it leaves the bow?

1 answer:

Nady [450]2 years ago

8 0

Answer:

v = 38.73 m/s

Explanation:

Given

Extension of the bow, x = 50 cm = 0.5 m

Force of the arrow, F = 150 N

Mass of the arrow, m = 50 g = 0.05 kg

speed of arrow, v = ? m/s

We start by finding the spring constant

Remember, F = kx, so

k = F/x

k = 150 / 0.5

k = 300 N/m

the potential energy if the bow when pulled back is

E = 1/2kx²

E = 1/2 * 300 * 0.5²

E = 0.5 * 300 * 0.25

E = 37.5 J

The speed of the arrow will now be found by using the law of conservation of energy

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m, on substituting, we have

v² = (300 * 0.5²) / 0.05

v² = 75 / 0.05

v² = 1500

v = √1500

v = 38.73 m/s

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Since both forces are the same,

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The question is missing some parts. Here is the complete question.

An ideal gas is contained in a vessel at 300K. The temperature of the gas is then increased to 900K.

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(ii) (c) a factor of $\sqrt{3}$ ;

(iii) (c) a factor of $\sqrt{3}$ ;

(iv) (c) a factor of $\sqrt{3}$ ;

(v) (e) a factor of 3;

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For 300K:

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$V_{rms2}=30\sqrt{3}\sqrt{\frac{k_{B}}{m} }$

Comparing both veolcities:

$\frac{V_{rms2}}{V_{rms1}}= (30\sqrt{3}\sqrt{\frac{k_{B}}{m} }) .\frac{1}{30} \sqrt{\frac{m}{k_{B}} }$

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