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30 June, 00:17

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

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  1. 30 June, 00:27
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    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
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