Question

2.) A person is pushing a 750 kg car a distance of 30m with a constant pushing force of 550 N that is parallel to the ground. If the car is accelerated from rest to a speed of 1.5 m/s, find the coefficient of kinetic friction between the tires and the road. Use the Work kinetic energy theorem.

Answer 1

solw Guven , mass of car &m) = 750 kg distance cover (d) = 30m. pushing force parallel to ground (s) 2 550 x Initial speeding o final speed) - 1.5 m/s. change in Kinetic energy (K. E change) = Imut – zmu = mto - 1x 750x(15) - 843-75 joule Let the coefficient of kinetic friction be us So, frictional force (FuF mg x where og is gravitations acceleration = 9.8m/5 Then the resultant force that causes displacement I is F - Fp - Fx = (550 mg xM ) so the work done by the resultant fonce w= Fad from work energy theorem , we know that, work done = change in K.E. so Exa = 843-75 & (550- qux) 30 = 843-75 – mg x = 843.75 – 550 – ngu – – 521.875 ma - 521.875 750 *9.8 = 0.071 Therefore, the coefficient of kinetic friction is 0.071 30