Average force physics question?

Question

An automobile traveling at a speed of 30 m/s applies its brakes and comes to a stop in 4.9 s. If the automobile has a mass of 0.8 " × 10^3 kg, what is the average horizontal force exerted on it during braking? Assume the road is level.

Answer 1

u = 30 m/s
v = 0 m/s
t = 4.9 s
Use the 1st equation of motion :
a = (v - u) / t = 6.12 m/s^2

m = 0.8 * 10^3 kg

F = ma = 4.9 * 10^3 N

Hope this helps.

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Answer 2

I suspect the above answers are correct...up to a point. But I question the use of the term "average force." Average is defined as Sum(each value)/Number of data points. That we can find in any introductory statistics text. [See source.]

Thus, in order to have an average force, we are likely to have more than one data point N(f) >= 1, each with a discrete force f value. Then Favg = Sum(f)/N(f) is the average force by definition.

But, when we invoke the SUVAT motion equations (S = Ut + 1/2 At^2) and Newt's F = MA, we have but one data point N(F) = 1, with a value F. In this case, Favg = F/N(F) = F/1 = F. Which means the "average" is really the force itself and that force is applied continuously throughout the time t.

In truth, we cannot accumulate force like so-many raindrops in a bucket. Force exists only so long as there is acceleration of a mass. For example, while bouncing a ball, there is force generated only at each rebound off the floor (and off the hand if you're a purist). But adding each individual force has no meaning as the forces are not accumulable.

If each bounce generated f = 1 pound of force, what does ten bounces generate = 10f = F = 10 pounds? No, not ten pounds, that has no physical meaninng. For example, F = mA and A = F/m = 10f/m = 10ma/m = 10 a, the ball certainly is not accelerating ten times faster (the 10 a) after bouncing the ball ten times.

As a consequence Sum(f) has no meaning in the physical world and consequently we cannot have an "average force;" at least one that is not the force F itself with N(F) = 1. So, what I am suggesting for clarity and consistency, is that the question should have read "what is the horizontal force...." without the "average."

[Point of clarification...you can add forces, but they are the sums of concurrently active forces that interact. For example, a net force f = W - B; where W is a weight force and B is a buoyancy force, is a legitimate sum of forces acting concurrently on a mass floating/sinking in a liquid.]

Lest you think I'm picking nits here, I've recently seen answers suggesting an average force could be found by adding up the individual forces of impacting bullets. They actually summed n bullet masses (m) to give M = nm = Sum(m). And then did this F = Sum(f) = Sum(ma) = Sum(m)a = M dv/dt = nm dv/dt; where dv/dt = a the acceleration of each bullet when it strikes the target. In other words, what they did was add up the bullet masses and lump them together like a cannon ball of mass M. F would = M dv/dt if all that M were packed into a single cannonball, but not when M = nm.

Answer 3

with the aid of fact the vehicle starts at 30 m/s then ends a nil m/sec the fee is -30/m/sec (meaning the vehicle is decelerating no longer accelerating). Dividing the fee via the time required for the vehicle to stop = -30 m/sec / 4.9 sec = -6.12 m/sec^2. employing the equation F=ma the rigidity exerted to stop the vehicle = ma = (.8 x 10^3 kg) x (-6.12 m/sec^2) = -4896 kg m/sec ^2 or -4896 N.

Answer 4

Prasant 123 @ yahoomail answred correctly

Answer 5

you can plug in the numbers,

a = v/t

F = m*a