Question

a truck of mass 1800 kg moving with a speed of 54km/h applied brake and stopped with an uniform negative acceleration at a distance of 200m.Calculate the force applied by the brake of the truck and the work done before stopping

Answer 1

mass = 1800 kg

u = 54 km / h = 15 m / s

v = 0 m /s

s = 200 m

We have to calculate the retardation and work done.

so we use the eq v²- u² = 2as

i.e.   0² - 15² =  2 * a * 200

      - 225 = 400 * a

       a = -225 / 400

          = - 0.56 m/s²

retardation is -0.56 m/s²

Work done = Fs = ma * s

                   = 1800 * ( - 0.56 ) * 200

                   = - 201.6 kJ