Question

Determine from the following diagram a. The spring constant of the bowb. The Elastic Potential Energy storedc. What will be the initial velocity of the arrow once the bow is released?

Answer 1

F= 300 N

d = 0.6m

m = 30g = 0.03 kg

• a.

Apply Hooke's law formula:

F = k x

k= F/ x

Where:

k= spring constant

F= force = 300N

x = distance = 0.6 m

k = 300N/0.6m = 500N/m

• b.

PE = 1/2 k (x)^2

PE = 1/2 (500 N/m) (0.6)^2 = 90 J

• c.

KE = 1/2 m v^2

KE = kinetic energy = PE = potential energy

v = √KE/ (1/2*m) = √2KE/m = √ (2* 90 / 0.03 ) = 77.46 m/s

Answers:

a = 500 N/m

b= 90 J

C = 77.46 m/s