Question

A machine carries a 33.6 kg package from an initial position of r0 = (0.502 + 0.751 + 0.207) m at t0 = 0 s to a final position of r1 = (7.82 + 2.17 + 7.44) m at t1 = 11.9 s. The constant force applied by the machine on the package is F = (21.5 + 42.5 + 63.5) N.

Answer 1

Answer:

The work done on the package by the machine's force is 676.94 J.

Explanation:

Given that,

Mass of package = 33.6 kg

Initial position $r_{0}=(0.502i+0.751j+0.207k)\ m$

Final position $r_{1}=(7.82i+2.17j+7.44k)\ m$

Final time = 11.9 s

Force $F=(21.5i+42.5j+63.5k)$

Suppose we need to find the work done on the package by the machine's force

We need to calculate the displacement

Using formula of displacement

$d=r_{1}-r_{0}$

Put the value into the formula

$d=(7.82i+2.17j+7.44k)-(0.502i+0.751j+0.207k)$

$d=7.318i+1.419j+7.233k$

We need to calculate the work done

Using formula of work done

$W=F\dotc d$

$W=(21.5i+42.5j+63.5k)\dotc(7.318i+1.419j+7.233k)$

$W=676.94\ J$

Hence, The work done on the package by the machine's force is 676.94 J.