the question is a 570kg block of concrete has to be moved, so a bulldozer is brought in. the Bulldozer causes the block to to acelerate from rest to 0.84m/s over a distance of 1.26m. the coefficient of friction between the boulder and the ground is 0.76, what force is the bulldozer aplying?
Force the bulldozer is applying =Fa mass=m=570 kg Acceleration of gravity = g = 9.8 m/s^2 force of friction = f =coefficient of friction(mu) * normal reaction coefficient of friction(mu)=0.76 f=(mu)mg f=0.76*570*9.8 =42.4536 N_____________(1) As friction opposes tendency of motion, 'Fa' and 'f' are in opposite direction. The resultant force 'F'= 'Fa' - 'f' As per Newton's second law, resultant force F=ma 'Fa' - 'f' =ma As initial velocity u=0, v^2 - u^2 =2as gives a= v^/2s 'Fa' - 'f'=ma =mv^/2s Fa = f + m v^2 / 2s Fa = (mu)mg + m v^2 / 2s Fa =m [ (mu)g + v^2 / 2s ] Fa = 570 [0.76*9.8 + (0.84)^2 / ( 2*1.26) ] =4404.96 N the bulldozer applying 4404.96 N force on the block ______________________________________...
You can solve this problem using kinematics equations, but I think conservation of energy is fewer steps and simpler math: First, find the force due to friction (g*m*coef) 9.81 * 570 * 0.76 = 4250 N Next, find the total kinetic energy imparted to the block after it has moved (0.5*m*v^2) 0.5*570*0.84^2 = 201J Finally, plug this into the work=force*distance equation, remembering that the net force will be the force of the bulldozer minus the frictional force: (Fdozer - 4250) * 1.26 = 201 Solve for Fdozer F=4410 N
