A spacecraft with mass 1.5x105 kg is planning to land on a planet with gravitati

Question

A spacecraft with mass 1.5x105 kg is planning to land on a planet with gravitational acceleration of g = 6.5m/s2 . When it is 260 m above the planet’s surface, it shuts down its engine and plunges in free fall.

a) Assuming an initial vertical velocity of zero and no air resistance, what is the velocity of the spacecraft when it hits the ground?

b) As the spacecraft touches the ground, it lands on top of a hill with a slope of 45 degrees and begins sliding down for 220 m. Assume that all kinetic energy of the spacecraft is converted into movement down the hill. Assuming a coefficient of friction of 0.24, what is the velocity of the spacecraft at the bottom of the hill?

Explanation / Answer

a) Applying , vf^2 - vi^2 = 2ad

v^2 - 0^2 = 2(6.5)(260)

v = 58.14 m/s

b) perpenndicular to the hill,

N - mgcos45 = 0

N = mg cos45

and friction force, f = ukN = 0.24 mg cos45

along the hill,

mgsin45 - 0.24mgcos45 = ma

a = 6.5(sin45 - 0.24cos45) = 3.5 m/s^2

using vf^2 - vi^2 = 2ad

v^2 - 58.14^2 = 2(220)(3.5)

v = 70.12 m/s

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