Question.
A constant retarding force of $50 \mathrm{~N}$ is applied to a body of mass $20 \mathrm{~kg}$ moving initially with a speed of $15 \mathrm{~ms}^{-1}$. How long does the body take to stop?
A constant retarding force of $50 \mathrm{~N}$ is applied to a body of mass $20 \mathrm{~kg}$ moving initially with a speed of $15 \mathrm{~ms}^{-1}$. How long does the body take to stop?
solution:
Retarding force, $F=-50 \mathrm{~N}$
Mass of the body, $m=20 \mathrm{~kg}$
Initial velocity of the body, $u=15 \mathrm{~m} / \mathrm{s}$
Final velocity of the body, $v=0$
Using Newton's second law of motion, the acceleration (a) produced in the body can be calculated as:
$F=m a$
$-50=20 \times a$
$\therefore a=\frac{-50}{20}=-2.5 \mathrm{~m} / \mathrm{s}^{2}$
Using the first equation of motion, the time $(t)$ taken by the body to come to rest can be calculated as:
$v=u+a t$
$\therefore t=\frac{-u}{a}=\frac{-15}{-2.5}=6 \mathrm{~s}$
Retarding force, $F=-50 \mathrm{~N}$
Mass of the body, $m=20 \mathrm{~kg}$
Initial velocity of the body, $u=15 \mathrm{~m} / \mathrm{s}$
Final velocity of the body, $v=0$
Using Newton's second law of motion, the acceleration (a) produced in the body can be calculated as:
$F=m a$
$-50=20 \times a$
$\therefore a=\frac{-50}{20}=-2.5 \mathrm{~m} / \mathrm{s}^{2}$
Using the first equation of motion, the time $(t)$ taken by the body to come to rest can be calculated as:
$v=u+a t$
$\therefore t=\frac{-u}{a}=\frac{-15}{-2.5}=6 \mathrm{~s}$