Physics, asked by amanverma6789, 8 months ago

a constant retarding force of 50 n is applied to a body of mass 10 kg moving initially with a speed of 10m/s the body comes to rest after -

Answers

Answered by sparshraghav123

Explanation:

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${\mathfrak{\underline{\red{\huge{\fbox{Answer = 2 second}}}}}}$

Attachments:

Answered by Mysterioushine

GIVEN :-

Magnitude of constant retarding force acting on the body is 50N
Mass of the body = 10 kg
Initial speed = 10m/s

TO FIND :-

Time taken by the body to come to rest

SOLUTION :-

Relation between Force , mass and acceleration is given by ,

$\large {\underline {\boxed {\bigstar {\red {\sf{ \: a = \frac{F}{m} }}}}}}$

$\implies \sf \: a = - \frac{50}{10} \\ \\ \implies \sf \: a = - \frac{5 \cancel{0}}{1 \cancel{0}} \\ \\ \implies {\underline {\boxed { \pink{\sf \: a = - 5 \: m {s}^{ - 2} }}}}$

Here negative sign of acceleration indicates the retardation in the body

we have ,

a = -5m/s²
u = 10m/s
v = 0 m/s

From first equation of motion ,

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

Where ,

v is final velocity
u is initial velocity
a is acceleration
t is time

$\implies \sf \: 0 = 10 + ( - 5)(t) \\ \\ \implies \sf \:0 = 10 + ( - 5t) \\ \\ \implies \sf \:0 = 10 - 5t \\ \\ \implies \sf \: - 10 = - 5t \\ \\ \implies \sf \cancel{ - }10 = \cancel{ - }5t \\ \\ \implies \sf \: 10 = 5t \\ \\ \implies \sf \: t = \frac{10}{5} \\ \\ \implies {\underline {\boxed {\blue{\sf {t = 2 \: s}}}}}$

∴ The time taken by the body to come to rest is 2 sec

ADDITIONAL INFO :-

◉ Second equation of motion is given by ,

$\large{\underline{\boxed{\bigstar{\red{\sf{v^2-u^2=2as}}}}}}$