I don't know how to do this. Working on #2

Physics

1 answer:

Mashcka [7]4 years ago

5 0

-- You know how much work the bow does on the arrow.

   Work = (force) x (distance) = (65 N) x (0.9 m)

-- THAT's the energy the arrow got from the bow, so
THAT's the kinetic energy the arrow has as it leaves the bow.

   KE = (1/2) (m) (speed)²

   so (1/2) (0.075 kg) (speed)² = (65 N) x (0.9 m)

The speed is the only thing left unknown in this mess.
I'm sure you can handle it from here, and find the speed.

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The 45-g arrow is launched so that it hits and embeds in a 1.40 kg block. The block hangs from strings. After the arrow joins th

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Question: How fast was the arrow moving before it joined the block?

Answer:

The arrow was moving at 15.9 m/s.

Explanation:

The law of conservation of energy says that the kinetic energy of the arrow must be converted into the potential energy of the block and arrow after it they join:

$\dfrac{1}{2}m_av^2 = (m_b+m_a)\Delta Hg$

where $m_a$ is the mass of the arrow, $m_b$ is the mass of the block, $\Delta H$ of the change in height of the block after the collision, and $v$ is the velocity of the arrow before it hit the block.