A man of mass 75.5 kg walks down the aisle of an airplane at a speed of 1.50 m/s in the forward direction while the plane moves at a speed of 305 m/s relative to the earth.
(a) Find the man's kinetic energy relative to the plane.
(b) Find the man's kinetic energy relative to the earth.
the speed relative to the plane is 1.50m/s find 1/2 m v^2
the speed relative to the ground is 304.5m/s find 1/2 mv^2
Thanks so much!
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To find the man's kinetic energy, we can use the formula:
Kinetic Energy (KE) = (1/2) * mass * velocity^2.
(a) To find the man's kinetic energy relative to the plane, we need to consider his motion relative to the plane. Since the man is walking down the aisle, his velocity relative to the plane is the same as his walking speed, which is 1.50 m/s forward.
Using the formula, we have:
KE (relative to plane) = (1/2) * mass * velocity^2
= (1/2) * 75.5 kg * (1.50 m/s)^2
= 84.79 Joules.
Therefore, the man's kinetic energy relative to the plane is approximately 84.79 Joules.
(b) To find the man's kinetic energy relative to the earth, we need to consider his motion relative to the earth. Since the plane is moving forward at a speed of 305 m/s, we need to add the velocity of the plane to the man's walking velocity.
Relative velocity of the man with respect to the earth = velocity of the plane + velocity of the man
= 305 m/s + 1.50 m/s
= 306.5 m/s.
Using the formula, we have:
KE (relative to earth) = (1/2) * mass * velocity^2
= (1/2) * 75.5 kg * (306.5 m/s)^2
= 3.59 * 10^6 Joules.
Therefore, the man's kinetic energy relative to the earth is approximately 3.59 * 10^6 Joules.