Question

ᴀ ᴄᴏɴꜱᴛᴀɴᴛ ʀᴇᴛᴀʀᴅɪɴɢ ꜰᴏʀᴄᴇ ᴏꜰ 50 ɴ ɪꜱ ᴀᴘᴘʟɪᴇᴅ ᴛᴏ ᴀ ʙᴏᴅʏ ᴏꜰ ᴍᴀꜱꜱ 20 ᴋɢ ᴍᴏᴠɪɴɢ ɪɴɪᴛɪᴀʟʟʏ ᴡɪᴛʜ ᴀ ꜱᴘᴇᴇᴅ ᴏꜰ 15 ᴍ/ꜱ. ᴛʜᴇ ᴛɪᴍᴇ ᴛᴀᴋᴇɴ ʙʏ ᴛʜᴇ ʙᴏᴅʏ ᴛᴏ ꜱᴛᴏᴘ ɪꜱ:

Answer 1

$\begin{gathered}\begin{gathered} \bf\large\underline\blue{Given:—} \\ \cal☛︎ Force, F = 50 N \\ \cal ☛︎ Mass, m = 20 kg \\ \cal☛︎ Initial \: velocity, u = 15 m/s \\ \\ \\ \bf\large\underline\blue{To\:find:—} \\ \cal \: How long \: does \: the \: body \: take \: to \: stop ? \\ \\ \\ \bf\large\underline\blue{Solution:—} \\ \tt \: From \: Newton's \: second \: law,we \: know \: that \\ \boxed{\bf{\red{★ F=ma}}} \\ ↦50 = 20 × a m/s \\ ↦a = 2.5 m/s \\ \\ \tt Now, \: as \: per \: first \: equation \: of \: motion:- \\ \boxed{\bf{\green{★ v=u+at}}} \\ ↦0 = 15 + (-2.5)×t \\ ↦t = 15/2.5 \\ \\ \boxed{\bf{\purple{t=6\:s}}} \\ \bf \: \small{Hence, the time taken by the body to stop is 6 seconds.}\end{gathered}\end{gathered}$

Answer 2

Answer:

F = - 50 N ( -- indicates retardation)

Mass of the body, m = 20 kg

Initial velocity of the body,

u= 15 m/s

Final velocity of the body, v= 0

UsingNewton's second law of motion, the acceleration (a)

produced in the body can be calculated as:

F = ma

$- 50 = 20 \times a$

$∴ \: a = \frac{ - 50}{20} = - 2.5$

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

v= u + at

$∴ \: t = \frac{ - u}{a}$

$= \frac{ - 15}{2.5}$

$∴ \: t = 6sec$