Question

A constant retarding force of 50N is applied to a body of mass 20kg moving initially with a speed of 15 $ms^{-1}$. How long does the body take to stop?

Answer 1

HI..

$F = - 50 \: N \: \\ m = 20 \: kg \: \\ u = 15 \: m {s}^{ - 1} \\v = 0 \: m {s}^{ - 1} \\ t = \: ? sec$

$F = m( \frac{v - u}{t} ) \\ \\ - 50 = 20( \frac{0 - 15}{t} ) \\ \\ \frac{ - 50}{20} = \frac{ - 15}{t} \\ \\ \frac{ - 5}{2} = \frac{ - 15}{t} \\ \\ t = \frac{ - 15 \times 2}{ - 5 } \\ \\ t = 6 \: sec$

HOPE THIS HELPS.

Answer 2

Explanation:

Retarding force, F = –50 N

Mass of the body, m = 20 kg

Initial velocity of the body, u = 15 m/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 = ma

–50 = 20 × a

∴ a = -50/20 = -2.5 ms-2

Using the first equation of motion, the time (t) taken by the body to come to rest can be calculated as:

v = u + at

∴ t = -u / a = -15 / -2.5 = 6 s