Question

A track of mass 1800 kg moving with a speed of 54 kg per hour.when brakes are applied it stops with uniform negative acceleration at a distance of 200 m. calculate the force applied by the brakes of the truck and the work done before stopping.

Answer 1

The initial velocity of the truck is, u=54km/hr u=54×518m/s u=15m/s Now, after applying brakes, the truck comes into rest position. thus, the final velocity of the truck will bev=0. The acceleration of the truck will be found by using equation of motion as, v2−u2=2as On substituting the values we get, (0)2−(15m/s)2=2a×200m−(15m/s)2=a×400ma=−225m2/s2400ma=−0.56m/s2 Negative sign shows that the retardation takes place when brakes are applied. Now, using Newton's second law of motion the retarding force will be calculated as, F=ma on substituting the values we get, F=(1800kg)×(−0.56m/s2)F=−1012.5N Thus, the magnitude of retarding force will be,F=1012.5N Now, the work done is defined as the product of force and the displacement, W=Fs On substituting the values we get, W=1012.5N×200mW=202500N.m