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 move 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. -6865.3 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 that 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 = 9.81m/86 m/s^2) 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.

Explanation / Answer

Applying Work - energy theorme,

net work done = change in KE

- 0.720 Favg = 0 - (71 x 11.8^2 / 2)

Favh = 71 x 11.8^2 / (2 x 0.720)

Favg = 6865.31 N ......Ans

(Here size of force is asked that means magnitude so negative sign is not used.)

a = - Favg / m = - 6865.31 / 71

= - -96.69

a ( in g ) = - 96.69 / 9.81 = - 9.87 ..........Ans

Don't use m/s^2 in Box . a in terms of g is pure number or unitless.

{ Ans: - 9.87 }

Favg = 71 x 11.8^2 / (2 x 0.210)

= 23538.2 N ......Ans

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