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 58.00 N/m, is 1.240 J. Find by how much is the spring is compressed.

A 0.090 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 1.70 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.

Explanation / Answer

A) Elastic Strain Energy = (1/2)kx^2

where,

k is the spring constant;

x is the extension
Now, to work out the extension;

x = sqrt(2E/k)

x = sqrt((2* 1.240)/ 58)

x = 0.206m

It is a form of energy so the units are obviously Joules we can work it out that they are Joules our units in the equation are Nm^-1 * m^2 which simplifies to Nm which using the formula Work = Force * Distance and given Work has unit Joules then Newtons*Metres = Joules

Therefore, x = 0.206m

B) I guess we are talking about the same dart gun. If so, and if the mass of the spring is negligable and it delivers all its energy to the dart then the energy of the dart is the same 1.240J you had before. So, as it goes up the kinetic energy will all convert to potential energy at the height where it stops. mgh = 1/2mv^2 = 1.18J and you can solve for h, h = 1.240J/(mg)

h = 1.240/(0.090*9.81)

h = 1.3948m

C) 1.240 J = (1/2)mv2
v = sqrt(2*1.240 (kg m2/s2) / 0.090 kg) = 5.249 m/s

D)   y = (1/2)gt^2

t = sqrt(2y/g) = sqrt(2*1.70 / 9.81) = 0.3465 s

Now, Horizontal Distance = Velocity(v) * Time(t)

Horizontal Distance = (5.249 m/s)(0.3465 s) = 1.818755m

All the best :)

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