Answer: 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