A projectile with mass of 167 kg is launched straight up from the Earth\'s surfa

Question

A projectile with mass of 167 kg is launched straight up from the Earth's surface with an initial speed vi. What magnitude of vi enables the projectile to just reach a maximum height of 3.0RE, measured from the center of the Earth? Ignore air friction as the projectile goes through the Earth's atmosphere.

_____?_____ km/s


#2) A 2.5-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spring is attached to a wall, as shown. The initial height of the block is 0.42 m above the lowest part of the slide and the spring constant is 485 N/m.

(a) What is the block's speed when it is at a height of 0.25 m above the base of the slide?
_____?_____ (The answer is not 2.2)

Explanation / Answer

a) Mass of the projectile m = 167 kg The height of the projectile reached is h = 3R As the projectile is projected vertically the final velocity is zero The kinetic energy of the projectile at the surface is K1 = 0.5mvi2 The gravitational potential energy at the surface is U1 = -GmM/R The kinetic energy of the projectile at the maximum height is K2 = 0 The gravitational potential energy at the maximum height is U2 = -GmM/3R According to the conseration of energy 0.5mvi2 - GmM/R = -GmM/3R vi = 4GM/3R vi = [4(6.67*10-11 Nm2/kg2)(5.98*1024 kg)/3(6.37*106 m)] vi = 9.137 km/s b) The mass of the block is m = 2.5 kg The initial height of the block h1 = 0.42 m The final height of the block is h2 = 0.25 m Let v be the final speed of the block The initial potential energy U1 = mgh1    The initial kinetic energy K1 = 0 The final potential energy U2 = mgh2 The final kinetic energy K2 = 0.5mv2 According to the conseration of energy mgh1 = mgh2 + 0.5mv2 v = 2g(h1-h2) v = [2(9.8 m/s2)(0.42 m - 0.25 m)] v = 1.825 m/s b) The mass of the block is m = 2.5 kg The initial height of the block h1 = 0.42 m The final height of the block is h2 = 0.25 m Let v be the final speed of the block The initial potential energy U1 = mgh1    The initial kinetic energy K1 = 0 The final potential energy U2 = mgh2 The final kinetic energy K2 = 0.5mv2 According to the conseration of energy mgh1 = mgh2 + 0.5mv2 v = 2g(h1-h2) v = [2(9.8 m/s2)(0.42 m - 0.25 m)] v = 1.825 m/s The initial kinetic energy K1 = 0 The final potential energy U2 = mgh2 The final kinetic energy K2 = 0.5mv2 According to the conseration of energy mgh1 = mgh2 + 0.5mv2 v = 2g(h1-h2) v = [2(9.8 m/s2)(0.42 m - 0.25 m)] v = 1.825 m/s The final potential energy U2 = mgh2 The final kinetic energy K2 = 0.5mv2 According to the conseration of energy mgh1 = mgh2 + 0.5mv2 v = 2g(h1-h2) v = [2(9.8 m/s2)(0.42 m - 0.25 m)] v = 1.825 m/s v = 2g(h1-h2) v = [2(9.8 m/s2)(0.42 m - 0.25 m)] v = 1.825 m/s

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