Braking force on a 1000 kg car stopping in 10 seconds

[galoshes]

I have an answer to this question and then a question:

A car with a mass of 1,000kg moves at 20 m/s. What is the braking force needed to bring the car to a halt in 10s.

I used m2v2-m1v1 and came up with -20,000 kg(m/s)
then -20,000kg(m/s)/10s

I came up with -2,000 N

Is this right? Does the negative mean slowing down?
Am I even on the right path?

 

[Corneo]

I agree with you answer. I believe the negative sign is correctly in place because say that car is traveling 20 m/s to the right taken as the positive direction, and we know acceleration is going to be negative. From Newton 2nd law F=ma, F will be in the same direction of a.

 

[galoshes]

thanks for the response. I hope I am right too, it's for a grade.

thanks again

 

[KnowledgeIsPower]

I have an answer to this question and then a question:

A car with a mass of 1,000kg moves at 20 m/s. What is the braking force needed to bring the car to a halt in 10s.

I used m2v2-m1v1 and came up with -20,000 kg(m/s)
then -20,000kg(m/s)/10s

I came up with -2,000 N

Is this right? Does the negative mean slowing down?
Am I even on the right path?

u = 20, v = 0 , t=10, a =?

v = u + at
0 = 20 + 10a
-20=10a
a=-20/10
a=-2m/s/s.

F=ma
F=1000x2
F=2000N, opposing the motion of the car.

 

[thermodynamicaldude]

yeah...knowledge of power...you calculations look fine to me...as force is a vector, so a negative value should simply mean that it is opposite to the motion of the car.