Question

a girl having a mass of 35 kg sits on a trolley og mass 5 kg. the trolley is given an initial velocity of 4m/s by applying a force. the trolley comes to rest after travelling a distance of 16m. how much work is done on the trolley? how much work is done by the girl?

Answer 1

U(intial velocity)= 4ms-1

v(final velocity)= 0ms-1

distance= 16m

. .
 

according to newton's 3rd law of motion.. 

v2= u2+ 2as

= (0)2= (4)2 +2a*16m

= 0 = 16 + 32a

= 16 = -32a

= a= -16/32

= a= -1/2

........

v= u +at

= t= v- u /a

= t= 0 -4 /-1/2

= t= 4*2

= t= 8 sec..

..

work done by trolley..

mass= 35kg+ 5kg= 40kg

force= m*a= 40kg*1/2

= 20N

...

work done= FS

= 20N*16m

= 320Nm..

..

work done by Girl.= FS

mass=5kg

force= 5kg*1/2

= 2.5N

..

work done= 2.5N*16m

= 40Nm ans..