Question

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

$\huge \tt \underline{Solution}$

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

$<marquee behavior = alternate><b><u> hope its helps$

Answer 2

Answer:

• $\textsf{6 seconds}$

Explanation:

$\star\:{\underline{\underline{\tt{\red{GIVEN}}}}}:-$

• $\sf\purple{Force \ (F) = 50N}$
• $\sf\green{Mass \ (m) = 20kg}$
• $\sf\purple{Initial \ velocity \ (u) = 15m/s}$
• $\sf\green{Final \ velocity \ (v) = 0}$

$\star\:{\underline{\underline{\tt{\orange{TO\: FIND }}}}}:-$

• $\sf\purple{Time \ taken \ to \ stop}$

$\star\:{\underline{\underline{\tt{\blue{SOLUTION}}}}}:-$

$\red{\underline{\boxed{\sf{Force = Mass \times Acceleration}}}}$

$\dag\:\underline{\mathfrak{\purple{Substituting \: the \: values}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\orange{ 50 = 20 \times a}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\blue{ a = \dfrac{50}{20}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\orange{ a = 2.5m/s^{2}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\blue{a = -2.5m/s^{2}}\:\:\:\sf\gray{(Due\: to \:retardation\: acceleration\: of \:body\: becomes \:negative.)}$

$\red{\underline{\boxed{\sf{v = u + at}}}}$

$\dag\:\underline{\mathfrak{\purple{Substituting \: the \: values}}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\orange{ 0 = 15 + (-2.5) \times t}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\blue{ 2.5t = 15}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\orange{ t = \dfrac{15}{2.5}}$

$\tiny\qquad\quad\rightarrow\:\:\:\:\sf\blue{ t = 6\:seconds}$

${\red{\underline{\boxed{\tt{\pink{Time\: taken\: to\: stop \:a\: body = 6\: seconds}}}}}}$