Question

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

Answer 1

Answer:

= 27 / 10 : 6.3 / 10 : 9 / 1 .

= 27 / 10 × 10 = 27 .

= 63 / 10 × 10 = 63.

= 9 / 1 × 10 .

= 90 .

H.C.F of 27, 63 & 9 = 9 .

Answer 2

$\begin{gathered} \bf{ \pmb{ \gray { \underline{GIVEN }}}} \\ \tt{ Mass \: of \: a \: body \: → \: 50kg }\\ \tt \: Velocity \: of \: body \: → \: 54km/hr \\ \\ \bf{ \pmb{ \gray{ \underline{ TO \: \: FIND }}}}\\ \tt \: \blue{Work \: done \: by \: body} \\ \\ \bf{ \pmb{ \gray{ \underline{SOLUTION}}}} \\ \small{ \sf \: According \: to \: Work \: Energy \: Theorem :−} \\ \\ \begin{gathered}\sf\red{ \boxed{ \bf Work \ Done = \Delta Energy_{Kinetic\ energy} } }\\\\\rm \small\: Work \ Done = \dfrac{1}{2}mv^2- \dfrac{1}{2}mu^2 \\\\\rm \small \: Work \ Done = \dfrac{1}{2}m ( v^2 - u^2 ) \\\\\rm \small Work \ Done = \dfrac{1}{2}\times 50kg [ (0m/s)^2 - (54km/hr)^2 ] \\\\\rm \small Work \ Done = 25kg \bigg\lgroup 0m^2/s^2 - \bigg(54 \times \dfrac{5}{18}m/s.\bigg)^2 \bigg\rgroup \\\\\rm \small \: Work \ Done = 25kg [ 0m^2/s^2 - (15m/s)^2 ] \\\\\rm \small \: Work \ Done = 25kg \times - 225 m^2/s^2 \\\\ \large \boxed{\blue{\mathbb {WORK \ DONE = -5625 \ JOULES .} }}\end{gathered} \end{gathered}$