Asked by sad
A 38 kg box with 300 J of Kinetic Energy is half way down a ramp with a velocity of 9 m/s. What is the remaining Potential energy of the box?
Total Energy = PE + KE
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Answered by
GPT-5 mini
AI
There's a contradiction in the numbers you gave. Compute the kinetic energy from the velocity:
KE = 1/2 m v^2 = 1/2 (38 kg)(9 m/s)^2 = 19·81 = 1539 J.
So if v = 9 m/s, the KE is 1539 J (not 300 J). On a frictionless ramp "halfway down" means half the initial gravitational potential energy has been converted to KE, so at that point PE = KE. Therefore the remaining potential energy is 1539 J.
(If instead the KE really were 300 J, the speed would be sqrt(2·300/38) ≈ 3.98 m/s, and at halfway the remaining PE would be 300 J.)
KE = 1/2 m v^2 = 1/2 (38 kg)(9 m/s)^2 = 19·81 = 1539 J.
So if v = 9 m/s, the KE is 1539 J (not 300 J). On a frictionless ramp "halfway down" means half the initial gravitational potential energy has been converted to KE, so at that point PE = KE. Therefore the remaining potential energy is 1539 J.
(If instead the KE really were 300 J, the speed would be sqrt(2·300/38) ≈ 3.98 m/s, and at halfway the remaining PE would be 300 J.)
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