Question

The potential energy stored in the compressed spring of a dart gun, with a spring constant of 36.00 N/m, is 1.440 J. Find by how much is the spring is compressed. A 0.070 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 4.30 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.

Answer 1

Energy stored Ue = (1/2)*k*x^2

1.44 = (1/2)*36*x^2

the spring is compressed by x = 0.2828 m

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along vertcical

at the maximum vertical distance potential energy E = m*g*h

from energy conservation

E = Ue

m*g*h = u

0.07*9.81*h = 1.44

vertical distance h = 2.097 m

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after firing horizontally

potential of spring = kinetic energy of dart

Ue = (1/2)*m*v^2

1.44 = (1/2)*0.07*v^2

v = 6.414 m/s

HORZIONTAL

PROJECTILE

along vertical
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v0y = 0

acceleration ay = -g = -9.8 m/s^2

initial position y0 = h

final position y = 0

from equation of motion

y-y0 = vy*T + 0.5*ay*T^2

-h = -(1/2)*g*T^2

T = sqrt(2h/g)

along horizontal
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initial velocity v0x = v

acceleration ax = 0

initial position = xo = 0

final position = x = ?

displacement = x - x0

from equation of motion

x - x0 = v0x*T+ 0.5*ax*T^2

x - 0 = v*T

x = v*sqrt(2h/g)

x = 6.414*sqrt(2*4.3/9.8)

x = 6 m <<<<----------------answer