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REY [17]
3 years ago
6

Calculate how fast the ball would be moving at the instant it leaves the projectile launcher of the spring is compressed by 3.75

cm. Use a value of k = 500 N/m for the spring constant, 10 g for the mass of the ball, and 75 g for the effective mass of the ball holder. Show your work.
Physics
1 answer:
Mnenie [13.5K]3 years ago
6 0

Answer:

V = 8.34m/s

Explanation:

Given that

1/2ke^2 = 1/2mv^2 ......1

Where e = 3.75cm = (3.75/100)m

e = 0.0375m

K = 500 N/m

m = 10g = 10/1000

= 0.01kg

Substitute the values into equation 1

0.5×500×(0.0375)^2 = 0.5×0.01×v^2

250×0.001395 = 0.005v^2

0.348 = 0.005v^2

v^2 = 0.348/0.005

v^2 = 69.6

V = √69.6

V = 8.34m/s

The ball launches at the speed of V = 8.34m/s

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