a. The spring is compressed by 0.174 m.
b. The vertical distance the dart travels from its position when the spring is compressed to its highest position is 9.6 m.
c. The horizontal velocity of the dart at that time is 13.74 m/s.
d. The horizontal distance from the equilibrium position at which the dart hits the ground is 19.236 m.
What is potential energy?
The energy by virtue of its position is called the potential energy.
a. Given is the potential energy stored in the compressed spring of a dart gun, with a spring constant of 62.00 N/m, is 0.940 J.
PE of spring = 1/2 kx²
Put the values, we get
P.E = 0.940 = 1/2 x 62 x²
x = 0.174 m
Thus, the spring is compressed by 0.174 m.
b. Given is a 0.010 kg dart is fired straight up.
The vertical height is find out by
0.940 J = (0.010 kg) (9.8 m/s²) h
h = 9.6 m
Thus, the vertical distance the dart travels from its position is 9.6 m
c. From the conservation of energy principle, total mechanical energy is conserved.
1/2 mv² =mgh
v = √2gh
Plug the values, we get
v = √2 x 9.8x 9.6
v = 13.74 m/s
Thus, the horizontal velocity is 13.74 m/s.
d. Time that dart spends in air, t = √2h/g
t = √(2x9.6)/9.81
t = 1.4 s
The horizontal distance from the equilibrium position at which the dart hits the ground.
Horizontal distance = (Velocity on x direction) x time
Horizontal distance = 13.74 m/s x 1.4s
Horizontal distance = 19.236 m
Thus, the horizontal distance is 19.236 m.
Learn more about potential energy.
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