An object with mass 120 kg moved in outer space. When it was at location < 13, -

Question

An object with mass 120 kg moved in outer space. When it was at location < 13, -29, -4 > its speed was 5 m/s. A single constant force < 220, 380, -180 > N acted on the object while the object moved from location < 13, -29, -4 > m to location < 17, -20, -7 > m. Then a different single constant force < 150, 210, 230 > N acted on the object while the object moved from location < 17, -20, -7 > m to location < 23, -27, -2 > m. What is the speed of the object at this final location?

final speed = ??? m/s

Explanation / Answer

Given that

Then a different single constant force < 150, 210, 230 > N

acted on the object while the object moved from location < 17, -20, -7 > m to location < 23, -27, -2 > m.

The displacement is given by <23-17,-27+20,-2+7> =<6,-7,5>

The work done is given by W =Force*displacement

=< 150, 210, 230 ><6,-7,5>=580J

Now applying the work energy theorem

W =(1/2)mv2

580J =0.5*120*v2

then the speed of the object at this final location is given by

v =3.11m/s

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