A space ship has four thrusters positioned on the top and bottom, and left and right as shown below. The thrusters can be operat
ed independently or together to help the ship navigate in all directions.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
Initially, the Space Probe is floating towards the East, as shown below, with a velocity, v. The pilot then turns on thruster #2.
Select one:
a.
Space ship will have a velocity to the West and will be speeding up.
b.
Space ship will have a velocity to the East and will be speeding up.
c.
Space ship will have a velocity to the East and will be slowing down.
d.
Space ship will have a velocity to the West and will be slowing down.
e.
Ship experiences no change in motion.
1 answer:
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Answer:
4.15 m/s
Explanation:
As the total energy must be conserved (neglecting air resistance) the change in gravitational potential energy, must be equal to the change in kinetic energy:
ΔE = ΔK + ΔU =0
If we take as a zero reference level for the gravitational potential energy, the height of the swing seat above the ground, (which is equal to 0.21 m), we can find the initial gravitational energy, considering the height of the point where the seat is released, regarding this point:
h₀ = 1.09 m -0.21 m = 0.88 m
⇒ U₀ = m*g*h₀ = 400 N*0.88 m = 352 J
As Uf = 0, ΔU = Uf -U₀ = -352 J
As the swing starts from rest, K₀=0, so we can say:
ΔK = Kf = (1)
As ΔK = -ΔU ⇒ ΔK = 352 J (2)
From (1) and (2) we can solve for vf, as follows:
So, when the swing passes through its lowest position, Betty moves at 4.15 m/s.