Question

A body of mass 2 kg moves under a force of \[(2\widehat i + 3\widehat j + 5\widehat {k)}{\rm{ N}}\]. It starts from rest and was at the origin initially. After 4s, its new coordinates are (8,b,20). The value of b is _______. (Round off to the Nearest Integer)

Answer 1

Hint:Displacement is the shortest path from initial to final position. So to get displacement we simply subtract the initial position from the final position. Here the body starts with rest so initial velocity will be zero. The second equation of motion gives the displacement of a body under constant acceleration.

Formula used:
Newton’s second law of motion,
\[\overrightarrow F = m\overrightarrow a \]
Where F is force applied, m is the mass of a body and a is acceleration.
Second equation of motion, we have
\[S = ut + \dfrac{1}{2}a{t^2}\]
Where, S is the displacement, u is the initial velocity, t is time taken and a is acceleration.

Complete step by step solution:
Given mass of a body, m= 2 kg
A constant force applied, \[F = (2\widehat i + 3\widehat j + 5\widehat k{\rm{) N}}\]
Initial velocity, \[\overrightarrow u = 0\]
Time taken, t=0
By Newton’s second law of motion,
\[\overrightarrow F = m\overrightarrow a \]
\[\Rightarrow \overrightarrow a = \dfrac{{\overrightarrow F }}{m}\]
\[\begin{array}{l} \Rightarrow \overrightarrow a {\rm{ }} = \dfrac{{(2\widehat i + 3\widehat j + 5\widehat k)}}{2}\\ \Rightarrow \overrightarrow a = (\widehat i + \dfrac{3}{2}\widehat j + \dfrac{5}{2}\widehat k){\rm{ m/}}{{\rm{s}}^2}\end{array}\]

Initial position, \[\overrightarrow {{x_i}} = (0\widehat i + 0\widehat j + 0\widehat k{\rm{) m}}\]
After 4s
Final position, \[\overrightarrow {{x_f}} = (8\widehat i + b\widehat j + 20\widehat k{\rm{) m}}\]
Now Displacement,
\[\overrightarrow S = \overrightarrow {{x_f}} - \overrightarrow {{x_i}} \]
\[ \Rightarrow \overrightarrow S= (8\widehat i + b\widehat j + 20\widehat k{\rm{) }}\]-\[(0\widehat i + 0\widehat j + 0\widehat k{\rm{) }}\]
\[\Rightarrow \overrightarrow S = (8\widehat i + b\widehat j + 20\widehat k{\rm{) m}}\]

By second equation of motion, we have
\[\overrightarrow S = \overrightarrow u t + \dfrac{1}{2}\overrightarrow a {t^2}\]
Substituting the given values
\[8\widehat i + b\widehat j + 20\widehat k\]=\[0 \times 4\] +\[(\widehat i + \dfrac{3}{2}\widehat j + \dfrac{5}{2}\widehat k){\rm{ }} \times {{\rm{(4)}}^2}\]
\[\Rightarrow 8\widehat i + b\widehat j + 20\widehat k\]=\[8\widehat i + 12\widehat j + 20\widehat k\]
\[\Rightarrow b\widehat j\]=\[12\widehat j\]
\[\therefore {\rm{ b = 12}}\]

The value of b is 12.

Note: Newton’s second law of motion can state the time rate of change of momentum of a body is equal to both the direction and magnitude of the force applied to it. If a force is applied to a body that causes it to accelerate, the body that gives changes its velocity at a constant rate. A body causes acceleration in the direction of force applied.