What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential di
1 answer:
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Answer:
The correct option is;
Man doing most work
Box gaining the most energy
D Y Q
Explanation:
The given parameters of the question are;
The distance the box P is pushed by the Man X = 0
The force the Man X applies to the box P = x N
The distance the box Q is lifted by the Man Y = h > 0 meters
The minimum force the Man Y applies to the box Q = W, the weight of the box
Work done = Force × Distance
Energy gained = Potential energy + Kinetic Energy = (Mass × Gravity × Height = Weight × Height = W × h) + 1/2 × Mass × Velocity²
The final velocity of either box = 0 m/s (The boxes are at rest on the ground or the shelf)
Therefore, Kinetic energy = 0 J
The work done by Man X = 0 × x = 0 J
The energy gained by the box P = W × 0 = 0 J
The work done by Man Y = W × h = W·h J
The energy gained by the box P = W × h = W·h J
We have, the work done by the man Y = W·h J is more than the work done by the man X = 0 J
The energy gained by the box P = W·h J is more than the energy gained by the box Q = 0 J
Therefore, the correct option is D, Man doing the most work is Y, box gaining the most energy is Q.