Assume that the force of a bow on an arrow behaves like the spring force. In aiming the arrow, an archer pulls the drawstring back 60 cm from its rest position and holds it in position with a force of 166 N. If the mass of the arrow is 54 g and the "spring" is massless, what is the speed (in m/s) of the arrow immediately after it leaves the bow?

42.95m/s

Explanation:

The magnitude of the force of the spring is given as

kx = 166N

From the question, we know that x = 60cm = 0.6m

Then, the constant of the spring would be k = 166/0.6 = 276.7N/m

When the bow is pulled back 60cm, the potential energy of the arrow is.

PE = 1/2kx²

PE = 0.5 * 276.67 * 0.6²

PE = 49.8J

Finally, the speed of the arrow can be found by using principle of conservation of energy, that states that, the total energy of an isolated system remains constant and conserved over time. This means, energy is neither created nor destroyed.

1/2kx² = 1/2mv²

kx² = mv²

v² = kx²/m

v² = 276.67 * 0.6²/ 0.054

v² = 1844.5

v = 42.95m/s

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the answer is c.)

explanation: