If a body of mass $200\ g$ falls from a height $200\ m$ and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact ( $g = 10\ m/s^{2}$)

Question

Dr. Abubacker · Accepted Answer

(b) $\mathrm{\Delta}U = mgh = 0.2 	imes 10 	imes 200 = 400J$
$	herefore$ Gain in K.E. = decrease in P.E. $= 400\ J$

Q. If a body of mass $200\ g$ falls from a height $200\ m$ and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact ( $g = 10\ m/s^{2}$)

Explanation

Reviewed by Dr. Abubacker

Want more practice?