A set of crash tests consists of running a test car moving at a speed of 11.8 m/

Question

A set of crash tests consists of running a test car moving at a speed of 11.8 m/s (26.4 mi/hr) into a solid wall. Strapped securely in an advanced seat belt system, a 71.0 kg (157 lbs) dummy is found to move a distance of 0.720 m from the moment the car touches the wall to the time the car is stopped. Calculate the size of the average force which acts on the dummy during that time. -6.865 10 9 N The force that acts on the dummy has to do (negative) work on the dummy. It must slow the dummy and stop it. Thus the amount of work equals the kinetic energy of the dummy. From the equation of work done by a force we can calculate the force. Using the direction of motion as the positive direction, calculate the average acceleration of the dummy during that time (in g's, with 1g = 9.81 m/s^2). -9.86 lbs In a different car, the distance the dummy moves while being stopped is reduced from 0.720 m to 0.210 m, calculate the average force on the dummy as that car stops. -2.358 10 4N

Explanation / Answer

average force that acted on dummy during 0.72 m movement Favg = mass*avg accleration

For calculating accleration we have final velocity v = 0, intial velocity u = 11.8 m/s, distance s = 0.72 m

using v2 = u2 + 2aS

0 = 11.8*11.8 + 2*a*0.72

a = -96.7 m/s2

Avergae force Favg = -71*96.7 = -6865.3 N = -1543.38 lbs

2) average accleration a = -96.7 m/s2 = -96.7/9.81 = -9.86g m/s2

3) if distance travelled is s = 0.21

using energy equation work done = change in Kinetic energy

F.s = 0 - 1/2Mu2

F = -0.5*71*11.8*11.8/0.21 = -23568.2 N = 5298.34 lbs

Your answers seems to be fine. Maybe try to upload lbs values or without sign.

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