Question

A tuck of mass 1800kg is moving with a speed of 54 km/h when breaks are applied it stops with uniform negative acceleration at a distance of200 m. Calculate the force applied by the brakes of the truck and the work done before stopping.

Answer 1

m = 1800kg , u = 54km/h = 15m/s , v = 0, s = 200m
f = ? , w = ?

$v {}^{2} = u {}^{2} + 2as$

$(0) {}^{2} = (15) {}^{2} + 2 \times a \times 200$

$0 = 225 + 400a$

$- 225 \div 400 = a$

$a = 0.56ms {}^{ - 2}$

F = m × a

= 1800 × 0.56

=1008N

W = F × s

= 1008 × 200

= 201600Nm

Thanks