Question

The momenta of a body
in two perpendicular direction at any time t are given by $P_x=2t^2+6$
and $P_y=\frac{3t^2}{2}+3$. The force acting on the body at $t=2$ sec is

• A. 5 units
• B. 2 units
• C. 10 units
• D. 15 units

Answer 1

The correct option is C 10 units

Given,

$P_x=2t^2+6$

$P_y=\frac{3t^2}{2}+3$

From the newton second law, the force is the rate of change of momentum,

$F=\frac{dP}{dt}$

The force acting on an object along x-axis,

$F_x=\frac{d{P}_x}{dt}$

$F_x=4t$

$F_x{|}_{t=2s}=4\times 2=8units$

The force acting along y-axis,

$F_y=\frac{d{P}_y}{dt}$

$F_y=\frac{3}{2}\times 2t=3t$

$F_y{|}_{t=2sec}=3\times 2=6units$

The force acting on the object,

$\overrightarrow{F}=F_x\hat{i}+F_y\hat{j}$

$F=\sqrt{F_x^2+F_y^2}$

$F=\sqrt{(8{)}^2+(6{)}^2}=\sqrt{100}$

$F=10units$

The correct option is C.