An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=598kg, m2=796kg and m3=311kg have blocked a busy road. 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 0.100m/s2. Use the value found above fro FA to find the force f12 exerted by the first rock of mass 598kg on the middle rock of 769kg

Question

Aryanna Taylor

Physics

24 October, 13:08

An overnight rainstorm has caused a major roadblock. Three massive rocks of mass m1=598kg, m2=796kg and m3=311kg have blocked a busy road. 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 0.100m/s2. Use the value found above fro FA to find the force f12 exerted by the first rock of mass 598kg on the middle rock of 769kg

+4

Answers (1)

Know the Answer?

Maci Sampson · Answer 1

The forse F12 = 108N

Explanation:

The total mass of the three rocks = 598kg + 796kg + 311kg = 1678kg

Following the second law of Newton F = m*a

⇒ with F = the vector sum of the forces F on an object (in Newton)

⇒ with m = total mass (kg)

⇒ with a = the accelaration (m/s²)

F1 = m*a = 1678 kg * 0.100 m/s² = 167.8 N

The force needed to move the first rock on its own is: F = 598 kg * 0.100 m/s² = 59.8 N

The force F12 is between F1 and Fa: F12 = (167.8 - 59.8) = 108 N

To control we can calculate F23 = (796 + 311) * 0.1 m/s² = 108 N