A force F = F, i + F,j acts on a particle
that undergoes a displacement of 3 = s, i +
8 y 3 where F, = 10 N, Fy = -5 N, s, = 6 m,
and sy = 2 m.
Find the work done by the force on the particle.
Answer in units of J.
The work done by a force on a particle is given by the dot product of the force vector and the displacement vector:
Work = F · s
In this case, the force vector F is given as F = Fx, i + Fy, j = 10N, i + (-5N), j = 10i - 5j.
The displacement vector s is given as s = sx, i + sy, j = 6m, i + 2m, j = 6i + 2j.
Now we can calculate the dot product:
Work = (10i - 5j) · (6i + 2j)
= 10i · 6i + (-5j) · 6i + 10i · 2j + (-5j) · 2j
= 60i • i + (-30j • i) + 20i • j + (-10j • j)
= 60(1) + (-30)(0) + 20(0) + (-10)(-1)
= 60 - 0 + 0 + 10
= 70
Therefore, the work done by the force on the particle is 70 J.