Question

A mass is suspended from the roof of a car by a string. While the car has a constant acceleration $a$, the string makes an angle of ${60}^o$ with the vertical. If $g=10m/s^2$, the value of $a$ is:

• A. $10\times (3{)}^{1/2}m/s^2$
• B. ${\left(\frac{10}{3}\right)}^{1/2}m/s^2$
• C. $5m/s^2$
• D. $5(3{)}^{1/2}m/s^2$

Answer 1

The correct option is A $10\times (3{)}^{1/2}m/s^2$

$T\sin \theta =ma$
$T\cos \theta =mg$
where $T$ is the tension in the string and $\theta$ is the angle the string makes with the vertical.
$\therefore tan\theta =\frac{a}{g}\Rightarrow a=gtan\theta =10\times tan{60}^o$
$a=10\sqrt{3}=\left(10\right)(3{)}^{1/2}m/s^2$