Question

A motor car of mass 2000 kg is moving with a velocity of 18 kmph is brought to rest by the application of block in distance of 50m. The force of resistance

Answer 1

$m = 2000kg \\ u = \frac{18 \times 1000}{60 \times 60} =5m |s \\ v = 0 \: m|s \\ k.e1 = \frac{1}{2}m {u}^{2} \\ = \frac{1}{2} \times 2000 \times 25 \\ = 1000 \times 25 = 25000j \\ \\ k.e2 = \frac{1}{2}m {v}^{2} \\ = \frac{1}{2} \times 2000 \times {0}^{2} \\ = 0j \\ w = k.e2 - k.e1 \\ = - 25000j \\ w = fs \\ - 25000 = f \times 50 \\ force = \frac{ - 25000}{50} = - 500newton$