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

Question

a) The potential energy stored in the compressed spring of a dart gun, with a spring constant of 43.50 N/m, is 0.700 J. Find by how much is the spring is compressed.


b) 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.

c) The same dart is now fired horizontally from a height of 3.50 m. The dart remains in contact until the spring reaches its equilibrium position. Find the horizontal velocity of the dart at that time.


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

Explanation / Answer

a)

Potential energy stored, PE = 0.5*k*x^2 <-------- x = compression

So, x = sqrt(2*PE/k)   

So, x = sqrt(2*0.7/43.5) = 0.179 m <-------answer

b)

Let the height reached be h

So, by energy conservation,

the potential energy stored in the spring is converted to the gravitational potetnial energy at the top

So, 0.5*kx^2 = 0.7 = mgh

So, h = 0.7/(0.07*9.8) = 1.02 m <-----answer

c)

Again using energy conservation,

the initial Potential energy of spring is converted to Kinetic energy

So, 0.5*kx^2 = 0.7 = 0.5*mv^2

So, v = sqrt(2*0.7/0.07) = 4.47 m/s <--------answer

d)

Using the equation of motion,

s = ut +0.5*at^2

For vertical motion,

u = 0 <---- initial velocity is horizontal

a = 9.8 m/s2

s = vertical displacement = 3.5 m

So, 3.5 = 0.5*9.8*t^2

So, t = 0.845 s

Now, for horizontal motion,

Horizontal distance covered when it hits ground = v*t

= 4.47*0.845

= 3.78 m <--------answer

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