Question : Under the action of a force (10i-3j-6k) N, a body of mass 5 kg moves from position (6,5,-3) m to a position (10,-2,7) m. Calculate the work done (in J) by the force.
Doubt by Suhani
Solution :
Here
F = (10i-3j-6k) N
r1= (6i+5j-3k) m
r2 = (10i-2j+7k) m
s = r2-r1
s = (10i-2j+7k)-(6i+5j-3k)
s = 4i-7j+10k
Now,
s = r2-r1
s = (10i-2j+7k)-(6i+5j-3k)
s = 4i-7j+10k
Now,
w = F.S
w = (10i-3j-6k).(4i-7j+10k)
w = (10)(4)+(-3)(-7)+(-6)(10)
w = 40+21-60
w = 61-60
w = 1 Joule
w = (10i-3j-6k).(4i-7j+10k)
w = (10)(4)+(-3)(-7)+(-6)(10)
w = 40+21-60
w = 61-60
w = 1 Joule
Hence, the required work done by the force is 1 Joule.