Newton's third law &normal force

Question

A car on flat ground accelerates at 0.785 m/s^2 when a net force of 1050 N acts on it. How much normal force acts on the car?

Answer 1

Assuming the net force acting on the car is horizontal:

 

F = 1050 N

a = 0.785 m/s²

 

If there is no vertical acceleration, the normal force F_n = weight = mg

 

m = F/a = 1050/0.785 kg = 1337.6 kg

 

F_n = mg = (1337.6 kg)(9.8 m/s²) ≅ 13108 N

Answer 2

Edited for math error...

 

F_N = mg, where m=mass and g=9.81m/s²

 

First, you need to solve for mass

 

You know acceleration and net force, so you can use the equation

 

F = ma

 

1050 N = m (.785 m/s²)

 

Newtons = kg * m/s², so

 

m = 1050 / .785 kg

m = 1337.58 kg

 

Now plug this value for mass into the Normal Force equation

 

F_N = mg

F_N = 1337.58 kg * 9.81 m/s²

F_N = 13121.66 kg * m/s² = 13121.66 Newtons