Question

An object is moving with a speed of 1.0 m/s A force $F_1$ is required to stop it over a distance X.If the speed of the object increases to 3.0 m/s, a force $F_2$ is required to stop it over the same distance X. $F_1$ : $F_2$ will be

• A. 1 : 3
• B. 3 : 1
• C. 1 : 9
• D. 1 : 6

Answer 1

The correct option is C 1 : 9

Given,

At first event, speed $V_1=1m/s$ & Force $F_1$

At second event, speed$V_2=3m/s$ & Force $F_2$

Both cover same distance while deaccelerating to a distance $X$

Kinetic Energy = Work done by force

At first event, $\frac{1}{2}m{V_1}^2=F_{1}X......\left(1\right)$

At second event, $\frac{1}{2}m{V_2}^2=F_{2}X......\left(2\right)$

Divide equation (1) by (2)

$\frac{F_1}{F_2}=\frac{{V_1}^2}{{V_2}^2}=\frac{1}{3^2}=\frac{1}{9}$

Ratio of forces $F_1:F_2=1:9$