Question

An object with mass 70 kg moved in outer space. When it was at location <7, -35, -5> its speed was 5.0 m/s. A single constant force <210, 400, -150> N acted on the object while the object moved from location <7, -35, -5> m to location <11, -27, -8> m. Then a different single constant force <140, 250, 210> N acted on the object while the object moved from location <11, -27, -8> m to location <17, -31, -3> m. What is the speed of the object at this final location?

Answer 1

Work done is given by : W=. where is force vector and is displacement vector,i.e., work done is given by dot product of force and displacement.

For the first force, F=<210,400,-150> N,

                          displacement=final position vector-initial position vector=<11,-27,-8> - <7,-35,-5>=<4,8, - 3> m

                          So,work done=<210,400,-150> . <4,8, - 3>=210*4+400*8+(-150)*(-3)=4490 J

For the second force, F=<140,250,210> N,

                            displacement=final position vector-initial position vector=<17,-31,-3> - <11,-27,-8>=<6,-4, 5> m

                                So,work done=<140,250,210> . <6,-4,5>=140*6+250*(-4)+210*5=890 J

So,total work done=4490+890=5380 J

Now,kinetic energy=1/2m*v² where m is mass of the object and v is its velocity

Work energy theorem says,change in kinetic energy=work done

So,final kinetic energy-initial kinetic energy=5380

=>1/2(70)v²-1/2(70)(5)²=5380    where v is the final velocity.

=> 1/2(70)v²=5380+875=6255

=>v²=6255/35=178.7 =>v=13.37 m/s