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A box of mass $m$ slides down a frictionless inclined plane of length $L$ and vertical height $h .$ What is the change in its gravitational potential energy?(A) $-m g L$(B) $-m g h$(C) $-m g L / h$(D) $-m g h / L$

Physics 101 Mechanics

Chapter 5

Work, Energy, and Power

Work

Kinetic Energy

Energy Conservation

Rutgers, The State University of New Jersey

University of Michigan - Ann Arbor

Simon Fraser University

Lectures

04:16

In mathematics, a proof is…

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A box of mass $m$ slides d…

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A block of mass $m$ slides…

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A block leaves a frictionl…

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A 75.0-kg skier rides a 28…

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Hi. In the given problem, the box off Mass M is sliding down a friction less inclined plane. The land off this inclined plane has been given us l and its height hide. From there, this box starts dropping down. Starts sliding down is given as edge. So when this box goes to the bottom off this inclined plane, there will be a reduction in its gravitational potential Energy and gravitational potential. Energy depends only on the vertical height, and it has given us m g h. So when this box slides down the incline, the change in gravitational potential energy will be. We can say it. Delta U. The final gravitational potential energy at the ground level, which is zero because the height at this point with zero miners, the gravitational potential energy at the top. Most point off this inclined plane, which is M g h. So finally, the answer will be m g h minus. M g h. Hence we can see option be is correct. Thank you

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