Question

Calculate the work required to be done to stop

a car of mass 1200 kg moving at a velocity

80 km/h.​

Answer 1

Explanation:

GIVEN

$\sf \: Mass \: of \: car \: \: = 1200 \: kg$

$\sf \: Velocity \: of \: car \: = 80 \: km \setminus \: h \\ \\ = \sf \: 80 \times \frac{1000}{3600} \: m \setminus \: s \\ \\ = \sf \frac{200}{9} \: m \setminus \: s$

$\sf \: we \: know \: when \: an \: \: object \: \: is \: \: moving \: \\ \sf \: it \: \: posses \: kinetic \: energy \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: so \: to \: \: stop \: the \: car \: we \: \: put \: \: the \: same \: \\ \sf work \: done \: as \: kinetic \: \: energy \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\sf \: work \: done = kinetic \: energy\: \\ \sf \: Kinetic \: Energy = \frac{1}{2} m {v}^{2} \: \: \: \: \: \: \: \:$

$\sf \: Kinetic \: Energy = \frac{1}{2} \times (1200) \times ( \frac{200}{9} ) { }^{2} \\ \\ = \sf \: 600 \times \frac{200 \times 200}{9 \times 9} \\ \\ = \sf \: 296296.3 \: joule$

$\sf \: \: hence \: work \: done \: to \: \: stop \: the \: car \\ \sf \: is \: 296296.3 \: joule \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Answer 2

hope it helps

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