A set of crash tests consists of running a test car moving at a speed of 10.6 m/s (23.3 m/h) into a solid wall. Strapped securely in an advanced seat belt system, a 63.0 kg (138.6 lbs) dummy is found to move a distance of 0.600 m from the moment the car touches the wall to the time the car is stopped.
1) Calculate the size of the average force which acts on the dummy during that time.
in N) 5.899×103 N
2)Using the direction of motion as positive direction, calculate the average acceleration of the dummy during that time (in g's) (use 1g=9.8 m/s2).
-9.554

3)In a different car, the distance the dummy moves while being stopped is reduced from 0.600 m to 0.290 m calculate the average force on the dummy as that car stops.

so I'm done with 1 and 2 , now I can not do 3 ! *** help
(in N )??

If you solved 1, you should have no problem with 3. Using:



you know v_f = 0, v_i = 10.6 m/s, and d=0.29 m, so solve for a.

I got a = -193.7 m/s^2
F=-ma
m= W/g
F=-193.7*6.4
and I got a wrong answer!!

The mass of the dummy is 63 Kg -- you should not divide that mass by g. You have the value for acceleration right, so F = ma yields F = 63 Kg * (-193.7 m.s^2) = -12204 N.

It would have helped you to write out the units - then you would have seem that dividing the dummy's mass of 63 Kg by 9.8m/s^2 doesn't make any sense. Remember that in SI units mass is given in Kg and weight is in N (since weight is a force). So if you ever wanted to know how much a 63 Kg dummy weighs, the answer is 63 Kg * 9.8 m/s^2 = 617.4 N.

The answer is incorrect !
It's like an online task !
And it giving me that the answer is not right !

Please check that you are entering the correct number of significant digits. Also, while I wrote that the force is negative (meaning it opposes the direction of travel), perhaps the on-line system doesn't want the negative sign (just like your answer for #1). Finally, please verify that the problem as you wrote it here is correct - that is, that the car decelerates from 10.6 m/s to 0 m/s in 0.29m.

I got it :) thanks