The spring constant of a toy dart gun is 5000 N/m. To load the gun the spring is compressed 3 cm.?

Question

The 7 g dart, fired straight upward, reaches a maximum height of 24 m. Determine the energy dissipated by air friction during the dart's ascent.
Estimate the speed of the projectile when it returns to its starting point.

Answer 1

k = 5000 N/m
x = 0.030 m
m = 0.007 kg
h = 24 m
g = 9.81 N/kg
v = speed of dart

energy stored in the spring = 1/2 kx^2 = 2.25 J
gravitational potential energy at highest point = mgh = 1.65 J
difference = energy lost due to air friction = 0.60 J


if it loses the same amount on the way down,
total energy left = 1.05 J
this will be converted to kinetic energy as it falls

kinetic energy = 1/2mv^2

v^2 = 2 KE/m

v = 17.3 m/s

Answer 2

ok = 100kg/s^2 x = 0.1m h = 5m PEsi(potential capability of spring)preliminary = PEgf(potential capability by way of gravity)very final........equa (a million) first resolve for PEsi PEsi = (a million/2)ok(x)^2 = (a million/2)(100kg/s^2)(0.1m)^2 = 0.5kgm^2/s^2 kgm^2/s^2 is Joule then PEsi = 0.5J PEgf = mgh m = ? g = 9.8m/s^2 h = 5m PEsi = PEgf ......equa(a million) 0.5J = mgh 0.5J = m(9.8m/s^2)(5m) now resolve for m as you remeber J=kgm^2/s^2 m = (0.5kgm^2/s^2)/(9.8m/s^2)(5m) m^2 and s^2 all cancels m = 0.01kg is the mass of dart a) 0.5Joule of capability is saved in the compressed spring. b) 0.5Joule is the aptitude capability of the dart on the suitable of its flight. c) the mass of the dart is 0.01kg.