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Sheikh Adil Grade: 12
        

A body of mass 10 kg is being acted upon a force 3t^2 and an opposing constant force of 32N. the initial speed is 10m/s. Find velocity of the body after 5 sec.

6 years ago

Answers : (2)

vikas askiitian expert
510 Points
										

net force on particle  = (3t2-32)N


net force = ma


ma = 3t2-32


10a = 3t2-32


a = 1/10 (3t2-32)      


now a = dv/dt 


 dv/dt = 1/10 (3t2-32)


 dv = 1/10 (3t2-32)dt


vf - u = 1/10 (3(t3/3) - 32t)            lim 0 to 5


vf = u + 1/10 (t3-32t)


at t=5 & u =10m/s


vf = 10 - 3.5 = 6.5m/s


so velocity  of particle is 6.5m/s .....


approve my ans if u like it

6 years ago
Technical Hacks
29 Points
										 

net force on particle  = (3t2-32)N

 

net force = ma

 

ma = 3t2-32

 

10a = 3t2-32

 

a = 1/10 (3t2-32)      

 

now a = dv/dt 

 

 dv/dt = 1/10 (3t2-32)

 

 dv = 1/10 (3t2-32)dt

 

vf - u = 1/10 (3(t3/3) - 32t)            lim 0 to 5

 

vf = u + 1/10 (t3-32t)

 

at t=5 & u =10m/s

 

vf = 10 - 3.5 = 6.5m/s

 

so velocity  of particle is 6.5m/s .....

 

approve my ans if u like it

thanks highfive

one month ago
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