Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A spring of spring constant 5×10m-1 is stretched initially by 5 cm from the unstretched position. Then the work required to stretch it further by another 5 cm is
PhysicsWork Power EnergyNEET
Options:
  • A 12.50 N m
     
  • B 18.75 N m
     
  • C 25 N m
     
  • D 6.25 N m
     
Solution:
1072 Upvotes Verified Answer
The correct answer is: 18.75 N m
 
W1=12kx12=1×52×103×5×10-22=6.25 J

W2=12kx1+x22

=12×5×1035×10-2+5×10-22=25 J

Net work done =W2-W1

=25-6.25=18.75 J=18.75 m

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.