Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A spring of spring constant 5 × 103 N/m is stretched initially by 5cm from the unstretched position. Then the work required to stretch it further by another 5 cm isa) 12.50 N-mb) 18.75 N-mc) 25.00 N-md) 6.25 N-m
Here, W=1/2k(x22-x12) = 1/2×5×103((10/100)2-(5/100)2) =1/2×5×103/104(100-25) =18.75JHence the work required to stretch the spring further by another 5cm is 18.75 N-mThanks and regards
Work done by spring=1/2k(x22-x12)K=5×103N/mX2=10×10-2mX1=5+10-2mW=1/2×5×103(100-25)10-4W=18.75N-m
Find solution of question on doubt solving app Instasolv
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -