Question

under a force of 10 ICAP - 3 J cap + 6 kcap Newton a body of mass 5 kg is displaced from the position of 6 ICAP + 5 - 3 to the position of 10 ICAP - 2 J cap + 7 k cap calculate the work done​

Answer 1

Given-:

Force (F)= 10icap-3jcap+6kcap and a body of mass 5kg displaced from position 6icap+5jcap-3kcap to the position of 10icap-2jcap+7kcap

To Find-:

The Work Done

Solution-    We Know the Work done formula i.e.

              W= F.S ,  where W is the work done in Joule , F is the force in             Newton , S is the displacement in m

•      Now by putting all the values we get

       W = ( 10icap-3jcap+6kcap ). ( 10icap-2jcap+7kcap-( 6icap+5jcap-3kcap ))

      W=( 10icap-3jcap+6kcap ).(4icap-7jcap+10kcap)

   now find the dot product of these we will get

      W= 40+21+60= 121J

The total work done is 121J