Here\'s the question.. Two cars, A&B, are traveling with the same speed of40.0m/

Question

Here's the question..
Two cars, A&B, are traveling with the same speed of40.0m/s, each having started from rest. Car A has a mass of1.20x10^3kg, and car B has a mass of 2.00x10^3kg. Compared to thework required to bring car A up to speed, how much additional workis required to bring car B up to speed?
Any help would be greatly appreciated! The only hint I gotfrom my instructor was W=KEf-KEo= (1/2mv^2f) - (1/2mv^2o)
Two cars, A&B, are traveling with the same speed of40.0m/s, each having started from rest. Car A has a mass of1.20x10^3kg, and car B has a mass of 2.00x10^3kg. Compared to thework required to bring car A up to speed, how much additional workis required to bring car B up to speed?
Any help would be greatly appreciated! The only hint I gotfrom my instructor was W=KEf-KEo= (1/2mv^2f) - (1/2mv^2o)

Explanation / Answer

the initial kinetic energy of car A is K1i = (1/2)m1 * u1^2 the final kinetic energy of car B is K1f = (1/2)m1 * v1^2 the additional work required to bring car A up to speedis W = K1f - K1i = (1/2) * (m1 * v1^2 - m1 * u1^2) m1 = 1.20 * 10^3 kg u1 = 0 v1 = 40.0 m/s similarly, the initial kinetic energy of car B is K2i = (1/2)m2 * u2^2 the final kinetic energy of car B is K2f = (1/2)m2 * v2^2 the additional work required to bring car A up to speedis W = K2f - K2i = (1/2) * (m2 * v2^2 - m2 * u2^2) m2 = 2.00 * 10^3 kg u2 = 0 v2 = 40.0 m/s the initial kinetic energy of car B is K2i = (1/2)m2 * u2^2 the final kinetic energy of car B is K2f = (1/2)m2 * v2^2 the additional work required to bring car A up to speedis W = K2f - K2i = (1/2) * (m2 * v2^2 - m2 * u2^2) m2 = 2.00 * 10^3 kg u2 = 0 v2 = 40.0 m/s

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---