Question: 1. A force F = 20 + 10y acts on a particle in y-direction where F is in newton and y in meter. Work done

1. 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 :
(1) 30 J
(2) 5 J
(3) 25 J
(4) 20 J


2. A force acts on a 30 g particle in such a way that the position of the particle as a function of time is given by x = 3t – 4t + t , where x is in metres and t is in seconds. The work done during the first 4 second is :-
(1) 5.28 J
(2) 450 mJ
(3) 490 mJ
(4) 530 mJ


3. An object of mass 4kg is dropped from a height 20 m. It reaches the ground with speed 16 m/s. Work done on the body by air friction is : (g = 10 m/s )

(1) 288 J 

(2) –288 J

(3) 312 J 

(4) –312 J

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1 To find the work done by the force to move the particle from y 0 to y 1 m we need to calculate the integral of the force with respect to y over the ... View full answer

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