2. (8 points) Spring Landing. Use the Work-Energy Theorem whenever possible. A b

Question

2. (8 points) Spring Landing. Use the Work-Energy Theorem whenever possible. A ball with mass m 0.045 kg is dropped down onto a vertical spring from a height of H 0.9 m above the equilibrium point. What is the ball speed v as it hits the spring? The spring constant is k 240 N/m. What is the maximum distance d below the equilibrium point that the ball reaches, as it stops momentarily, before going back up? Suppose the ball sticks securely to the end of the spring. When it goes back up, by what distance h will it rise above the equilibrium point? The ball and spring go up and down for many cycles, slowing down until they come to rest at a new equilibrium point, a distance s below the old one. What is s? _m/s, d - (Dropping): v- (Rising): h (New equilibrium point): s 7t,

Explanation / Answer

xe = compression of the spring at equilibrium = ?

k = spring constant = 240 N/m

m = mass = 0.045 kg

Using equilibrium of force

k xe = mg

(240) xe = (0.045) (9.8)

xe = 0.00184 m

H = height above the equilibrium = 0.9 m

h = height dropped by the mass before hitting the spring = H - xe = 0.9 - 0.00184

Using conservation of energy

kinetic energy gained = potential energy lost

(0.5) m v2 = mgh

v = sqrt(2 gh)

v = sqrt(2 x 9.8 (0.9 - 0.00184 ) )

v = 4.2 m/s

d = compression of the spring below the equilibrium position

Using conservation of energy

Potential energy lost = spring potential energy gained

mg (H + d) = (0.5) k (xe + d)2

(0.045) (9.8) (0.9 + d) = (0.5) (240) (0.00184 + d)2

d = 0.0575 m = 5.75 cm

h = d = 0.0575 m = 5.75 cm

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