The potential energy stored in the compressed spring of a dart gun, with a sprin

Question

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 34.50 N/m, is 1.280 J. Find by how much is the spring is compressed.

A 0.190 kg dart is fired straight up. Find the vertical distance the dart travels from its position when the spring is compressed to its highest position.

The same dart is now fired horizontally from a height of 4.50 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.

Find the horizontal distance from the equilibrium position at which the dart hits the ground.

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Explanation / Answer

1.
PE = 1/2kx^2
x = sqrt(2PE/k) = sqrt (2 x 1.280 / 34.50)
x = 0.272 m

2.
PE = mgh
h = PE / mg = 1.280 / (0.190 x 9.8)
h = 0.69 m

3.
PEspring = KEdart
PEspring = 1/2mv^2
v = sqrt (2PE/ m) = sqrt (2 * 1.280 / 0.19)
v = 3.67 m/s

4.
time to fall comes from height h,
t = sqrt(2h/g) = sqrt (2 x 4.5 / 9.8)
t = 0.96 sec
now, d = vt = 3.67 * 0.96
d = 3.524 m

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