Question bank/JEE Main/Physics
JEE MainPhysicsModerate
A force given by the relation F = 8t, acts on a body of mass 2 kg,
initially at rest. Find the work done by this force on the body during
first 2 seconds of its motion ( in joule )
m = 2 kg F = 8t
2 sec. ( )
Show answer and solution
F = 8t
m[IMAGE] = 8t
[IMAGE] mv = 4t2
or m [IMAGE] = 4t2 ⇒ dx =[IMAGE] W =[IMAGE] =[IMAGE] = 64 J
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE
m[IMAGE] = 8t
[IMAGE] mv = 4t2
or m [IMAGE] = 4t2 ⇒ dx =[IMAGE] W =[IMAGE] =[IMAGE] = 64 J
TOPIC: WPE
SUB TOPIC: WORK DONE BY VARIABLE FORCE
Source: Grass-G / P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx
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