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It takes a force of 80.0 N to compress the spring of a toy popgun 0.200m to "load" a 0.180-kg ball. With what speed will the ball leave the gun?

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Below is the solution:

9.43 m/s 
F = -kx 
k = 400 N/m 
PE= 0.5kx^2 = 0.5mv^2 
solve, v=9.43 m/s

dotmanjohn

Answer:

v = 9.43 m/s

Explanation:

According to Hooke's law;

F = kx ------------- eqn 1

where F = spring force = 80 N

k = spring constant = ?

x = spring stretch = 0.200 m

substitute for F and x to find k

80 = 0.2k

\(k = \frac{80}{0.2} = 400 N/m\)

\(\frac{1}{2} kx^{2} = \frac{1}{2} mv^{2}\)

\(\frac{1}{2}X400X0.2^{2} = \frac{1}{2}X 0.18v^{2}\)

\(200 X 0.04 = 0.09v^2\)

\(8 = 0.09v^2\)

\(v^2 = \frac{8}{0.09}\)

\(v^2 = 88.89\)

\(v = \sqrt{88.89}\)

\(v = 9.43m/s\)

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