Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
JEE Main & Advanced
Physics
Work, Energy, Power & Collision / कार्य, ऊर्जा और शक्ति
Question Bank
Conservation of Energy and Momentum
question_answer
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 \[(g=10\,m/{{s}^{2}})\] [AFMC 1997]
A) 200 J
B) 400 J
C) 600 J
D) 900 J
Correct Answer:
B
Solution :
\[\Delta U=mgh=0.2\times 10\times 200\]= 400 J \ Gain in K.E. = decrease in P.E. = 400 J.