Answer:
(a) 46.94 J.
(b) 55.95 m/s
Explanation:
(a)
Potential Energy: This is the energy of a body, due to its position. The S.I unit of potential energy is Joules (J).
The formula of potential energy in a stretched spring is
Ep = 1/2ke² .......................... Equation 1
Where Ep = potential energy of the spring, k = Force constant of the spring, e = extension or compression.
Given: k = 425 N/m, e = 0.47 m.
Substitute into equation 1
Ep = 1/2(425×0.47²)
Ep = 46.94 J.
(b)
at the instant When the arrow leaves the bow, the potential energy of the arrow is converted kinetic energy of the bow.
I.e,
Ep = 1/2mv² ............. Equation 2
Where m = mass of the arrow, v = velocity of the arrow.
make v the subject of the equation
v = √(2Ep/m)............. Equation 3
Given: Ep = 46.94 J, m = 0.03 Kg.
Substitute into equation 3
v = √(2×46.96/0.03)
v = √(93.92/0.03)
v = √(3130.67)
v = 55.95 m/s
