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 47.50 N/m, is 0.540 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 1.90 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)

energy in spring

U = ½ kx2
x = sqrt(2*U/k)

x = sqrt(2*0.54/47.50)

x = 0.1507m
x = 0.15m

(b)
By energy conservation
The stored energy in the spring = 0.54J = gravitational potential energy at the top = m*g*h

h = 0.54/(m*g)

h = 0.54/(0.090*9.8)

h = 0.61m

c)

U spring = K dart
so

0.54J = ½ mv2

v = sqrt(2*0.54/0.090)

v = 3.46m/s

d)

The time to fall comes from y = ½ gt2
so

t = sqrt(2*y/g)

t = sqrt(2*1.90/9.80)

t = 0.62269s

Horizontal range = v*t

r = 3.46m/s*0.62269s

r = 2.157 m

r = 2.15 m

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