A body of mass 5 kg resting on a smooth horizontal plane, is acted upon by force 6i + 2j, 5i + 4j and 4i - j. Calculate the: (a) velocity of the body (b) Ma...
Question 1 Report
A body of mass 5 kg resting on a smooth horizontal plane, is acted upon by force 6i + 2j, 5i + 4j and 4i - j. Calculate the:
(a) velocity of the body
(b) Magnitude of its velocity after 4s.
Answer Details
A body of mass \(m=5\text{ kg}\) on a smooth horizontal plane (so no friction) is acted on by the forces \(6\mathbf{i}+2\mathbf{j},\ 5\mathbf{i}+4\mathbf{j},\ 4\mathbf{i}-\mathbf{j}\) (in newtons). Assume it starts from rest.
Resultant force.
\[\mathbf{F}=(6+5+4)\mathbf{i}+(2+4-1)\mathbf{j}=15\mathbf{i}+5\mathbf{j}\text{ N}.\]
Acceleration from Newton's second law \(\mathbf{a}=\dfrac{\mathbf{F}}{m}:\)
\[\mathbf{a}=\frac{15\mathbf{i}+5\mathbf{j}}{5}=3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\]
(a) Velocity after 4 s. Starting from rest, \(\mathbf{v}=\mathbf{a}t:\)
\[\mathbf{v}=(3\mathbf{i}+\mathbf{j})\times4=12\mathbf{i}+4\mathbf{j}\text{ m/s}.\]
(b) Magnitude of the velocity after 4 s.
\[|\mathbf{v}|=\sqrt{12^2+4^2}=\sqrt{144+16}=\sqrt{160}=4\sqrt{10}\approx12.65\text{ m/s}.\]
(Here part (a) is taken as the velocity vector after 4 s, obtained via the resultant force and acceleration \(3\mathbf{i}+\mathbf{j}\text{ m/s}^2.\))
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