Physics, asked by MiniDoraemon, 5 months ago

A spring of spring constant 5× 10³ N/m is stretched initially by 5 cm from the unstratched position . then the work required to stretch it further by another 5cm [AIEEE 2003] ​

Answers

Answered by avijitm88
3

Explanation:

Work done is change in potential energy

The spring is initially stretch to 5 cm and again further stretched by 5 cm and 

further stretched by 5 cm

w=p10−p5

=21×5×103[I(10010)2−(1005)2]

=21×5×103(10010+1005)(10010−1005)N−m

=21×5×103×10015×1005

=2.5×15×5×103×10−4

=75×2.5×10−1

=7.5×2.5N−m

=18.75N−m

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