Question

A constant retarding force of 50N is applied to a body of mass 20kg moving initially with speed of 20m/s .How long does the body take to stop?​

Answer 1

ANSWER:

• The time taken by the body to stop = 8 s.

GIVEN:

• A constant retarding force of 50N is applied to a body of mass 20kg.

• Initial speed = 20 m/s.

• Final speed = 0.

TO FIND:

• The time taken by the body to stop.

EXPLANATION:

$\bigstar \ \boxed{ \bold{ \large{ \red{Decceleration = - \dfrac{Force}{Mass}}}}} \\ \\ \\ \sf \longmapsto Decceleration = - \dfrac{50}{20}\\ \\ \\ \sf \longmapsto Decceleration = - \dfrac{5}{2} \\ \\ \\ \sf \longmapsto Decceleration = - 2.5 \ ms^{ - 2} \\ \\ \\ \bigstar \ \boxed{ \bold{ \large{ \orange{v = u + at}}}} \\ \\ \\ \sf \longmapsto 0 = 20 - 2.5(t) \\ \textsf{[ \ At last body comes to rest\ ]} \\ \\ \\ \sf 20 = 2.5t \\ \\ \\ \sf t = 8 \ s$

Hence the time taken by the body to stop = 8 s.

NOTE : Decceleration = Retardation = Negative acceleration.

EXTRA CALCULATION:

$\bigstar \ \boxed{ \bold{ \large{ \blue{v^2 - u^2 = 2as}}}} \\ \\ \\ \sf \longmapsto 0^2 - 20^2 = 2(-2.5)S\\ \\ \\ \sf \longmapsto 0 - 400= -5S \\ \\ \\ \sf \longmapsto 400=5S \\ \\ \\ \sf \longmapsto S = 80\ m \\ \\ \\ {\Huge{\bold{ (OR)}}}\\ \\ \\ \bigstar \ \boxed{ \bold{ \large{\purple{S=ut +\dfrac{1}{2}at^2}}}} \\ \\ \\ \sf \longmapsto S = 20(8) +\dfrac{1}{2}(-2.5)(8)^2 \\ \\ \\ \sf \longmapsto S = 160 +\dfrac{1}{2}(-2.5)64 \\ \\ \\ \sf \longmapsto S = 160 -32(2.5) \\ \\ \\ \sf \longmapsto S = 160-80\\ \\ \\ \sf \longmapsto S = 80\ m$

Hence the body travels 80 m before coming to rest.