Physics. (a) How Much Work Is Done In Raising A 50 Kg Crate A Distance Of  M Above A Store Room Floor?

Physics. (a) How Much Work Is Done In Raising A 50 Kg Crate A Distance Of M Above A Store Room Floor?

Physics. (b) What Is The Change Of Potential Energy As A Result Of This Move? (c) How Much Kinetic Energy Will The Crate Have As It Falls And Hits The

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Physics : A How Much Work Is Done In Raising A 50 Kg Crate A Distance Of M Above A Store Room Floor

Physics. A How Much Work Is Done In Raising A 50 Kg Crate A Distance Of  M Above A Store Room Floor.

(b) What is the change of potential energy as a result of this move? (c) How much kinetic energy will the crate have as it falls and hits the floor? ~~~ sojo3 ~~~

Best Answer To Physics Question

(a) W=mgh = 50** = 735 J (b) delta Ep = 735 J (c) Ek = 735 J Doug

All Answers To Physics Questions

Answer 1

(a) W=mgh = 50** = 735 J (b) delta Ep = 735 J (c) Ek = 735 J Doug

Answer 2

a) You're raising a 50kg mass above the floor in a gravitational field with /s/s. So it means that the work done is .8m/s/s=735 joules b) the work done is converted to potential energy within the gravitational field, so the potential energy = work done = 735 joules c) likewise, just as the crate impacts the floor, that potential energy gets coverted in full to the kinetic energy such that the kinetic energy = potential energy at above the floor = 735 joules This sounds like your homework..so hope it helps!

Answer 3

i have no idea, i'm not a scale.

Answer 4

Potential energy is given by: U=m*g*h where m=object mass, g=acceleration due to gravity (-/s^2), and h=height of the object. Generally speaking, this problem is moving the mass from h1 to h2 (where h1=0 and h2=). Thus deltaU=U2-U1 deltaU=m*g*h2-m*g*h1 deltaU=m*g*(h2-h1) deltaU=m*g*(-0) deltaU=50*-* deltaU=-735 J Now, the work you did (force you applied) to the crate was in the same direction as its motion, so you know that the work done was positive. W=-U W=735 J Kinetic energy is always positive, and is equivalent to the change in potential on the way down. Thus KE=-U, or KE=735J The other posters have not attempted to reconcile their signs.

Answer 5

(a) Work done = Force x Distance ........................= m xg xh ..........................= 50 x x ..........................= 735 Joules. (b) Change in PE = Same as work done = 735 Joules (c) The whole PE is converted into KE = 735 Joules

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