Question

A force 10i-3j+6k Newton acts on a body of 5 kg and it displaces from 6i+5j-3kmetre to 10i-2j+7k metre.the work done is-

Answer 1

So, we know the starting displacement:
<6,5,-3> and the ending displacement: <10,-2,7>
We need to know how long the change in displacement is to calculate work. So, we need to use the distance formula.
\[\sqrt{(i_2-i_1)^2+(j_2-j_1)^2+(k_2-k_1)^2}\]
\[\sqrt{(10-6)^2+(-2-5)^2+(7--3)^2}\]
\[\sqrt{4^2+(-7)^2+10^2}\]
\[\sqrt{165}m\]
Now we need to find the magnitude of the force:
<10,-3,6>
\[\sqrt{10^2+(-3)^2+6^2}=\sqrt{145}N\]
Now we know the distance and the force applied, the eqn for work being:
\[W=F*d\]
\[W=\sqrt{145}\sqrt{165}=5\sqrt{957} J\approx154.6770J\]

Answer 2

sorry agreena the options are
a)100J b)0J
c)121J d)no of this
ans is 121 but I can't solved this
but I also get the same ans