Question

A constant retarding force of 50 \(\text N\) is applied to a body of mass 20 \(\text {kg}\) moving initially with a speed of 15 \(\text m \,\text s^{-1}\). How long does the body take to stop ?

Answer 1

Retarding force, \(\text F\) = –50 \(\text N\)
Mass of the body, \(\text m\) = 20 \(\text {kg}\) 
Initial velocity of the body, \(\text u\) = 15 \(\text m/\text s\) 
Final velocity of the body, \(\text v\) = 0 
Using Newton’s second law of motion, the acceleration (a) produced in the body can be calculated as:
\(\text F\) = \(\text {ma}\)
–50 = 20 × \(\text a\)
\(\therefore\)  \(\text a\) = \(\frac{-50}{20}\) = -2.5 \(\text m/\text s^2\)
Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:
\(\text v\) = \(\text u+\text {at}\)
∴ \(\text t\) = \(\frac{-u}{a}\) = \(\frac{-15}{-2.5}\) = 6 \(\text s\)