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

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

Step-by-step 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.