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Lina20 [59]
4 years ago
15

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 o

f 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.
Physics
1 answer:
Luda [366]4 years ago
7 0

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.

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