A question about Change in Momentum

AI Thread Summary
To bring a 1,000 kg car moving at 20 m/s to a halt in 10 seconds, a braking force of -2,000 N is required. The negative sign indicates that the force is acting in the opposite direction of the car's motion, confirming it is a deceleration. Using the formula F=ma, the acceleration calculated is -2 m/s², which aligns with the need for a force opposing the car's initial velocity. The calculations and reasoning presented are correct, affirming the understanding of momentum and forces in this context. This discussion highlights the importance of recognizing the directionality of forces in physics problems.
galoshes
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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?
 
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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.
 
thanks for the response. I hope I am right too, it's for a grade.

thanks again
 
galoshes said:
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.
 
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.
 
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