What average force is required to stop a 1400 kg car in 6.0 s if the car is traveling at 90 km/h ? express your answer to two significant figures and include the appropriate units. enter positive value if the direction of the force is in the direction of the initial velocity and negative value if the direction of the force is in the direction opposite to the initial velocity.

F=5833.3 N N

Explanation:

Newton's second law applied to the car

F= m*a Formula (1)

F: Force in Newtons (N)

m : mass in kg

a: acceleration ( m/s²)

kinematics car

vf= v₀ + a*t  Formula (2)

vf : final velocity (m/s)

v₀ : final velocity  (m/s)

a : acceleration  ( m/s²)

t : time t

Equivalences

1 km= 1000m

1 h = 3600 s

Data

m= 1000kg

v₀ = 90 km/h = 90*1000/3600 m/s = 25 m/s

vf= 0

t= 6 s

Problem Development

We calculate the acceleration replacing the data in the formula (2) :

0 =  25 + a*6

a= -25/6 = -4.16 m/s² ( The negative sign indicates that the car is braking)

We calculate the force is required to stop the car replacing the data  in the formula (1)

-F = 1400 kg*(-4.16 m/s²)

F=5833.3 N

the box does not move because you need to use a force greater than 300 n. you are only using 50 n. it needs to at least be 301 n for the box to move because then it is an unbalanced force and it will move the way that the greater force is pushing it. if you use only 300 n it will be a balanced force and will not move because the force you are exerting on the   box is the same as its force.

hope this !

im sorry if i went too in depth, i am going to include a link to a video on khan academy that will explain this.   hope that you like the video and that it explains more in detail to you.

that is the link.

total distance/total time

(85 + 82)/(5+3)

167/8

20 just over