An object with mass 80 kg moved in outer space. When it was at location <7, -34, -7> its speed was 14.0 m/s. A single constant force <200, 460, -150> N acted on the object while the object moved to location <12, -42, -11> m. What is the speed of the object at this final location?

Respuesta :

The speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

What is the speed of a body?

The speed of a body is the rate at which it covers the total distance is in the time taken.

If the speed of the body has direction, then it is known as the velocity. The velocity is the vector quantity.

The constant force applied on the object is  <200, 460, -150> N. It can be written in the vector form as,

[tex]\vec F=200\hat i+460\hat j-150\hat k[/tex]

The initial position vector of the object for the location <7, -34, -7> can be given as,

[tex]d_i=7\hat i-34\hat j-7\hat k[/tex]

The final position vector of the object is for the location <12, -42, -11> m can be given as,

[tex]d_f=12\hat i-42\hat j-11\hat k[/tex]

Thus, the total distance traveled by the constant force is,

[tex]d=(12\hat i-42\hat j-11\hat k)-(7\hat i-34\hat j-7\hat k)\\d=5\hat i-8\hat j-4\hat k[/tex]

The work done by a object is the product of force and displacement. Thus, work done on the object is,

[tex]W=\vec F.\vec d \\W=(200\hat i+460\hat j-150\hat k).(5\hat i-8\hat j-4\hat k)\\W=1000-3680+600\\W=-2080 \rm J[/tex]

Now, the work done on a object is equal to the kinetic energy of the body. Thus, work done on the object is,

[tex]W=\dfrac{1}{2}m(v^2_f-v^2_i)[/tex]

Here, v(f) is the final speed and v(i) is the initial speed of the object. As the mass of the object is 80 kg and initial speed is 14 m/s. Thus, put the values as,

[tex]-2080=\dfrac{1}{2}(80)(v^2_f-14^2)\\v_f=12\rm m/s[/tex]

Thus, the speed of the object with mass 80 kg and moved under a constant force at the final location is 12 m/s.

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