Question

A constant force = 2.00 n + 3.00 n acts on a 5.00 kg object as it moves in a straight line from the position1 = 1.00 m + 1.00 m to a position 2 = 4.00 m - 1.00 m . determine the work done by the force during this motion.

Answer 1

force acting on the object is given as

$F = 2.00\hat i + 3.00 \hat j$

initial and final positions are given as

$r_i = 1.00 \hat i + 1.00\hat j$

$r_f = 4.00 \hat i - 1.00\hat j$

now the displacement is given as

$d = r_f - r_i$

$d = (4.00 - 1.00)\hat i + (-1.00 - 1.00)\hat j$

$d = 3 \hat i - 2\hat j$

now the work done is given as

$W = F.d$

$W = (2.00\hat i + 3.00\hat j).(3\hat i - 2\hat j)$

$W = 0$

so there is no work done