Question

A car of mass $1000\text{kg}$ is moving with the speed $40\text{m/s}$ on a circular path of radius $400\text{m}$. If its speed is increasing at the rate of $3m{s}^{-2}$ the total force acting on the car is

• A. $3000\text{N}$
• B. $4000\text{N}$
• C. $5000\text{N}$
• D. $7000\text{N}$

Answer 1

The correct option is C $5000\text{N}$

We know that, the net force $F_{net}=m{a}_{net}$.
Here the car is moving in the circular path and hence there will be two type of acceleration, tangential and centripetal acceleration which are perpendicular to each other.

Hence the net acceleration
$a_{net}=\sqrt{(\frac{v^2}{rg}{)}^2+(a_T{)}^2}$, it is given that the speed is increasing at the rate of $3m/s^2$, so $a_T=3{\text{m/s}}^2$.

Hence the net force,
$F_{net}=m{a}_{net}=m\sqrt{(\frac{v^2}{rg}{)}^2+(a_T{)}^2}$
$\Rightarrow F_{net}=1000\sqrt{(\frac{1600}{400}{)}^2+3^2}={10}^3\sqrt{16+9}$
$F_{net}=5000\text{N}$

Hence option 'c' is the correct answer