A force F = 20 + 10y acts on a particle in y- direction where F is in ne...

A force F = 20 + 10y acts on a particle in y- direction where F is in newton and y in meter. Work done by this force to move the particle from y = 0 to y = 1 m is

AIPMT/NEET 2019 | W.P.E Q 2 | We had seen same qus in year 2005 and 2011( graph based ) so once again I advice you all please try to solve old Neet qus and if you are still not following my channel then please like , share and subscribe right now. Relation between force v/s distance is given , from which we can understand , the force is variable.

When Force is variable (either magnitude or direction) :
If the force applied on a body is changing its direction or magnitude or both , the force is said to be variable.
dW = Sd·F r r
Þ W = ò r Sd·F r ( F r cannot come out of integral since it is varying) = Area under force v/s distance curve. Simple calculation in this qus as compared to previous qus.

 

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