An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1 = 598 kg, m2 = 769 kg and m3 = 311 kg have blocked a busy road. The rocks are side by side blocking the road and lined up from left to right in order as m1, m2 and m3. The city calls a local contractor to use a bulldozer to clear the road. The bulldozer applies a constant force to m1 to slide the rocks off the road. Assuming the road is a flat frictionless surface and the rocks are all in contact, what force, FA, must be applied to m1 to slowly accelerate the group of rocks from the road at 0.400 m/s2?Fa=?????Use the value found above for FA to find the force, F12, exerted by the first rock of mass 598 kg on the middle rock of 769 kg. F12=????
It's really tempting to point out that if the road really is a flat frictionless surface, the bulldozer won't be able to apply much force :-) OK, the rocks are all in contact, so the total mass m = 598 + 769 + 311 = 1678 kg From Newton's Second Law, force = mass * acceleration Fa = 1678 * 0.400 = 671.2 N The force required to accelerate the first rock on its own to 0.400 m/s? would just be 598 * 0.400 = 239.2 N So force F12 is the difference between 239.2 N and Fa F12 = (671.2 - 239.2) = 432 N { Check: To accelerate the second and third rocks total mass (769 + 311) = 1080 kg by 0.400 m/s? requires a force = m * a = 1080 * 0.400 = 432 N. ok.}
