14) An 1100 kg car is traveling at 30 m/'s comes to a turn in the road, what radius of turn can the car take without slowing down if the coefficient of friction between the tires and the road is 0.7? 

15. Find the acceleration of the system if the hanging mass is 15kg and the mass on the table is 10kg. The coefficient of friction between the mass and the table is 0.3,

 

14) The normal force that the ground is pushing the car is given by N = mg = (1100 kg)·(9.8 m/s²) = 10,780 Newtons

The friction force can be found using Ff = µ·N = 0.7·(10,780 N) = 7,546 N

This net force will cause a radial acceleration (v²/r) within Newton's 2nd Law of Motion: Fnet = ma ==> Ff = m·v²/r. Thus,

7,546 N = (1100 kg)·(30 m/s)²/r ==> (solving for r) r = 131.1953 m ≈ 130 m (rounded to two significant figures)

15) The net force moving the system is the weight of the hanging mass, minus the friction force. 

Fnet = ma ==> (weight of the hanging mass) - (friction force) = (total moving mass)·a

==> (15 kg)·(9.8 m/s²) - (10 kg)·(9.8 m/s²)·0.3 = (15 kg + 10 kg)·a ==> a = 4.704 m/s² ≈ 4.7 m/s² (rounded to two significant figures)