Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Answer
Comment
Share
Q)

A person trying to lose weight by burning fat lifts a mass of $10\; kg$ upto a height of $1\; m \;1000\; times$. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies $3.8 \times 10^7 \;J$ of energy per kg which is converted to mechanical energy with a $20\%$ efficiency rate. Take $g=9.8 \;ms^{−2}$ :


( A ) $2.45 \times 10^{-3} kg$
( B ) $12.89 \times 10^{-3} kg$
( C ) $9.89 \times 10^{-3} kg$
( D ) $6.45 \times 10^{-3} kg$

1 Answer

Comment
A)
Efficiency =20÷100=0.2 potential energy=mgh=1×9.8×1000 Work done=energy =m×3.8×10_7 20÷100=10×9.8×1000÷(3.8×10_7×m) m=10×9. 8×1000×÷(0.2×3.8×10_7) =12.89×10_3
Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...