Question

calculate the work done required to be done to stop a car of 1500kg moving at a velocity of 60km/hr
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Answer 1

Answer:

w=F.s --------(1)

F= ma ---------(2)

v² = u²+2as. (final velocity, v= 0)

a=(v²-u²)/2s. a=-u²/2s

substitute in 2,

F= m(-u²/2s)

substitute in 1,

w= -1500 (u²/2s)× s. ( u=60×(5/18) = 16.67)

w= - 1500(16.6²/2)

Answer 2

SOLUTION :-

$\: m = \: 1500 \: kg$

➠ $v = \frac{60km}{h} \times \frac{5}{8} = \frac{50 \: {m}^{s} }{3}$

➠$ev = w = \frac{1}{2} {}^{3} \: mv {}^{2}$

➠ $\frac{1}{2} \times 1560 \times \frac{50}{3} {}^{2} = \frac{1}{2} \times 1500 \times \frac{50}{3} \times \frac{50}{3}$

➠ $\frac{250 \times 2500}{3} = \frac{625000}{3}$

Hence, proved