Question

A body of mass 2 kg is moving with a velocity of $\overline{V}=3\check{i}+4\hat{j}m/s$. A steady force $\overline{F}=\check{i}-2\hat{j}N$ begins to act on it. After four seconds, the body will be moving along.

• A. X-axis with a velocity of 2 m/s
• B. Y-axis with a velocity of 5 m/s
• C. X-axis with a velocity of 5 m/s
• D. Y-axis with a velocity of 2 m/s

Answer 1

The correct option is C X-axis with a velocity of 5 m/s

Given that F=i-2j, we can say that the acceleration will be a=0.5i-j since $a=\frac{F}{m}$and mass is given to be 2kg in this question. After this, we calculate the velocity after breaking initial velocity V=3i+4j into its components. Thus, in the x-direction, $Vx=u\left(x\right)+a\left(x\right)\times t=3+(0.5\times 4)=5$ m/s.

In the y direction,$Vy=u\left(y\right)+a\left(y\right)\times t=4-4=0$

Therefore we get answer C, which is 5 m/s in the x direction.