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Question: A body of mass \(5\;kg\) under the action of constant force \(\vec{F} = F_X\hat{i}+F_y\hat{j}\) has ...

A body of mass 5  kg5\;kg under the action of constant force F⃗=FXi^+Fyj^\vec{F} = F_X\hat{i}+F_y\hat{j} has a velocity at t=0  st=0\;s as v⃗=(6i^–2j^)  ms−1\vec{v}=(6\hat{i} – 2\hat{j})\;ms^{-1} and at t=10  st=10\;s as v⃗=6j^  ms−1\vec{v} = 6\hat{j}\;ms^{-1}. The force F⃗\vec{F} is:
A. (−3i^+4j^)  N(-3\hat{i}+4\hat{j})\;N
B. (−35i^+45j^)  N(-\dfrac{3}{5}\hat{i}+\dfrac{4}{5}\hat{j})\;N
C. (3i^−4j^)  N(3\hat{i}-4\hat{j})\;N
D. (35i^−45j^)  N(\dfrac{3}{5}\hat{i}-\dfrac{4}{5}\hat{j})\;N

Explanation

Solution

Recall that the force acting on a body executing kinematic motion can be given by the acceleration of motion experienced by the mass of the body. To this end, determine the acceleration of the body in question since we know that it is nothing but the rate of change of velocity. Following this, use the definition of force that we have established before to calculate the constant force that the body is subjected to.

Formula used:
Force ⇒F⃗=mΔv⃗t\Rightarrow \vec{F} = \dfrac{m\Delta \vec{v}}{t}

Complete step-by-step answer:
Let us begin by listing out all that is given to us.
We have the mass of the body m=5  kgm = 5\;kg
The initial velocity of the body is given to be u⃗=(6i^–2j^)  ms−1\vec{u}=(6\hat{i} – 2\hat{j})\;ms^{-1}
The final velocity of the body is given to be v⃗=6j^  ms−1\vec{v} = 6\hat{j}\;ms^{-1}, that the body achieves in time t=10  st=10\;s
Now, we know that the force acting on a body in kinematic motion can be given as:
F⃗=ma⃗\vec{F} = m\vec{a}
Since acceleration a⃗\vec{a} is the rate of change of velocity, its expression can be obtained from the kinematic equation of motion:
v⃗=u⃗+a⃗t⇒a⃗=v⃗−u⃗t\vec{v} = \vec{u}+\vec{a}t \Rightarrow \vec{a} = \dfrac{\vec{v}-\vec{u}}{t}
⇒F⃗=mΔv⃗t\Rightarrow \vec{F} = \dfrac{m\Delta \vec{v}}{t}
⇒F⃗=5×[(6j^)–(6i^–2j^)]10\Rightarrow \vec{F} = \dfrac{5 \times [(6\hat{j}) – (6\hat{i} – 2\hat{j})]}{10}
⇒F⃗=12.(−6i^+8j^)\Rightarrow \vec{F} = \dfrac{1}{2}.(-6\hat{i}+8\hat{j})
⇒F⃗=−3i^+4j^  N\Rightarrow \vec{F} = -3\hat{i} +4\hat{j}\;N
Therefore, the correct choice would be A. −3i^+4j^  N-3\hat{i} +4\hat{j}\;N

So, the correct answer is “Option A”.

Note: Remember to always constrain computations to directionally similar vectors only. We saw that the initial velocity was specified in i^\hat{i} and j^\hat{j} directions while the final velocity was specified only in the j^\hat{j} direction, which means that it possessed no velocity in the i^\hat{i} direction, which we take to be null. Thus while finding out the difference we consider only the corresponding velocity component in that specific direction of velocity. The same applies to any sort of calculations that need to be carried out in unit vector forms.