A seat belt holds a 70 kg crash test dummy in the car seat during a collision that slows the car from 90 km/hr to 0 km/hr. What is the approximate magnitude of the work done by the seat belt on the crash test dummy?A.2,000 JB.20,000 JC.40,000 JD.300,000 J
Question
A seat belt holds a 70 kg crash test dummy in the car seat during a collision that slows the car from 90 km/hr to 0 km/hr. What is the approximate magnitude of the work done by the seat belt on the crash test dummy?A.2,000 JB.20,000 JC.40,000 JD.300,000 J
Solution
To solve this problem, we first need to understand that the work done on the crash test dummy is equal to the change in kinetic energy of the dummy. The formula for kinetic energy is KE = 1/2 * m * v^2, where m is the mass and v is the velocity.
Step 1: Convert the velocity from km/hr to m/s. 90 km/hr = 90,000 m/3600 s = 25 m/s.
Step 2: Calculate the initial kinetic energy of the dummy before the collision. KE_initial = 1/2 * m * v^2 = 1/2 * 70 kg * (25 m/s)^2 = 21875 J.
Step 3: Calculate the final kinetic energy of the dummy after the collision. Since the final velocity is 0, the final kinetic energy is also 0.
Step 4: Calculate the work done by the seat belt. Work done = Change in kinetic energy = KE_final - KE_initial = 0 - 21875 J = -21875 J.
The negative sign indicates that the work is done against the motion of the dummy. However, since the question asks for the magnitude, we take the absolute value, which is 21875 J.
So, none of the options given in the question are correct. The work done by the seat belt on the crash test dummy is approximately 21875 J.
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