Question

calculate the work done required to stop a car of mass 50 kg moving witha velocity of 54 km/h​

Answer 1

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Given :- }}}$

$\sf\implies Mass \ of \ the \ car \ is \ 50kg .\\\sf\implies Initial\ Velocity \ is \ 54km/h.$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: To \ Find :- }}}$

$\sf\implies The \ work \ done \ to \ stop \ the \ car .$

$\Large\underline{\underline{\red{\sf \purple{\maltese}\:\: Answer :- }}}$

We will use Work Energy Theorem to find the work done , according to which work done is equal to the change in the Kinetic energy i.e. ∆ K.E. Since the car will be stopped the final Velocity will be 0m/s.

$\underline{\purple{\boldsymbol{ According \ to \ Work \ Energy \ Theorem :- }}}$

$\sf:\implies\pink{ Work \ Done = \Delta Energy_{Kinetic\ energy } }\\\\\sf:\implies Work \ Done = \dfrac{1}{2}mv^2- \dfrac{1}{2}mu^2 \\\\\sf:\implies Work \ Done = \dfrac{1}{2}m ( v^2 - u^2 ) \\\\\sf:\implies Work \ Done = \dfrac{1}{2}\times 50kg [ (0m/s)^2 - (54km/hr)^2 ] \\\\\sf:\implies Work \ Done = 25kg \bigg\lgroup 0m^2/s^2 - \bigg(54 \times \dfrac{5}{18}m/s.\bigg)^2 \bigg\rgroup \\\\\sf:\implies Work \ Done = 25kg [ 0m^2/s^2 - (15m/s)^2 ] \\\\\sf:\implies Work \ Done = 25kg \times - 225 m^2/s^2 \\\\\mathfrak:\implies \boxed{\pink{\mathfrak {Work \ Done = -5625 \ Joules .} }}$

$\underline{\blue{\sf \therefore Hence \ the \ work \ done \ is \ \textsf{\textbf{-5625 \ Joules }}. }}$