Answer:
a) the velocity is v=1.385 m/s
b) the ball has its maximum speed at 4.68 cm away from its compressed position
c) the maximum speed is 1.78 m/s
Explanation:
if we do an energy balance over the ball, the potencial energy given by the compressed spring is converted into kinetic energy and loss of energy due to friction, therefore
we can formulate this considering that the work of the friction force is equal to to the energy loss of the ball
W fr = - ΔE = - ΔU - ΔK = Ui - Uf + Kf - Ki
therefore
Ui + Ki = Uf + Kf + W fr
where U represents potencial energy of the compressed spring , K is the kinetic energy W fr is the work done by the friction force. i represents inicial state, and f final state.
since
U= 1/2 k x² , K= 1/2 m v² , W fr = F*L
X= compression length , L= horizontal distance covered
therefore
Ui + Ki = Uf + Kf + W fr
1/2 k xi² + 1/2 m vi² = 1/2 k x² + 1/2 m vf² + F*L
a) choosing our inicial state as the compressed state , the initial kinetic energy is Ki=0 and in the final state the ball is no longer pushed by the spring thus Uf=0
1/2 k X² + 0 = 0 + 1/2 m v² + F*L
1/2 m v² = 1/2 k X² - F*L
v = √[(k/m)x² -(2F/m)*L] = √[(8.07N/m/5.35*10^-3 Kg)*(-0.0508m)² -(2*0.033N/5.35*10^-3 Kg)*(0.16 m)] = 1.385 m/s
b) in any point x , and since L= d-(X-x) , d = distance where is no pushed by the spring.
1/2 k X² + 0 = 1/2 k x² + 1/2 m v² + F*[d-(X+x)]
1/2 m v² =1/2 k X²-1/2 k x² - F*[d-(X-x)] = (1/2 k X²+ F*X) - 1/2k x² - F*x + F*d
taking the derivative
dKf/dx = -kx - F = 0 → x = -F/k = -0.033N/8.07 N/m = -4.089*10^-3 m = -0.4cm
at x m = -0.4 cm the velocity is maximum
therefore is 5.08 cm-0.4 cm=4.68 cm away from the compressed position
c) the maximum speed is
1/2 m v max² = (1/2 k X²+ F*X) - 1/2k x m² - F*(x m) + 0
v =√[ (k/m) (X²-xm²) + (2F/m)(X-xm) ] = √[(8.07N/m/5.35*10^-3 Kg)*[(-0.0508m)² - (-0.004m)²] + (2*0.033N/5.35*10^-3 Kg)*(-0.0508m-(-0.004m)] = 1.78 m/s
