Question #234130

The force acting on a body of mass 10kg is (2i+j-k)N. If the body is initially at rest, determine the magnitude of velocity at the end of 20s


1
Expert's answer
2021-09-08T08:13:53-0400

Let the force be F=2i+jk\mathbf{F} = 2\mathbf{i} + \mathbf{j}-\mathbf{k}. Then, according to the Newton's law, the acceleration of the body is:


a=Fm\mathbf{a} = \dfrac{\mathbf{F}}{m}

where m=10kgm = 10kg is the mass of the body.

From the kinematic equation the velocity is expressed as follows (if body was initially at rest):


v=at\mathbf{v} = \mathbf{a}t

where t=20st = 20s is the time. Thus, obtain:


v=tmF=20s10kg(2i+jk)N==(4i+2j2k)m/s\mathbf{v} = \dfrac{t}{m}\mathbf{F} = \dfrac{20s}{10kg}\cdot(2\mathbf{i} + \mathbf{j}-\mathbf{k})N = \\ =(4\mathbf{i} + 2\mathbf{j}-2\mathbf{k})m/s

The magnitude is then:


v=42+22+(2)24.9m/sv = \sqrt{4^2 + 2^2 + (-2)^2} \approx 4.9m/s

Answer. 4.9 m/s.

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