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

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

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.