SPRING CONSTANT Q-In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35* angle.?
In a physics lab experiment, a compressed spring launches a 32 g metal ball at a 35degree angle. Compressing the spring 18 cm causes the ball to hit the floor 1.5m below the point at which it leaves the spring after traveling 6.0m horizontally.
What is the spring constant?
i have no idea how to solve this, it is due in an hour, can some1 please give me a step by step explanation and answer to this. thanks alot.
- Anonymous9 years agoBest Answer
x = Vo t cos Φ
t = x/(Vo cos Φ)
y = Vo t sin Φ + ½ g t²
y = Vo(x/(Vo cos Φ)) sin Φ + ½ g (x/(Vo cos Φ))²
y = x tan Φ + ½ g (x/(Vo cos Φ))²
-1.5 = 6 tan 35° + ½ (-9.8)(6/Vo)² sec² 35°
Vo = 6.79047 m/s
according conservation of energy,
ΔE = 0
½ m Vo² - ½ k (Δx)² = 0
½ (0.032)(6.79047)² - ½ k (0.18)² = 0
k = 45.5412 N/m