Asked by Anonymous
A 1.53×10
3
kg car accelerates uniformly from
rest to 13.6 m/s in 2.44 s.
What is the work done on the car in this
time interval?
Answer in units of J
3
kg car accelerates uniformly from
rest to 13.6 m/s in 2.44 s.
What is the work done on the car in this
time interval?
Answer in units of J
Answers
Answered by
Henry
Wc = m*g = 1530kg * 9.8N/kg = 14,994 N.
= Wt. of car.
a = (V-Vo)/t = (13.6-0)/2.44=5.57 m/s^2.
d = 0.5a*t^2 = 0.5*5.57*(2.44)^2=16.6 m.
Work = F*d = 14994*16.5 = 248780 Joules.
= Wt. of car.
a = (V-Vo)/t = (13.6-0)/2.44=5.57 m/s^2.
d = 0.5a*t^2 = 0.5*5.57*(2.44)^2=16.6 m.
Work = F*d = 14994*16.5 = 248780 Joules.
Answered by
Henry
Correction:
a = (V-Vo)/t = 13.6-0)/2.44=5.57 m/s^2.
F = m*a = 1530*5.57 = 8522.1 N.
d=0.5*a*t^2 = 0.5*5.57*(2.440^2=16.6 m.
Work = F*d = 8522.1*16.6 = 141467 Joules
a = (V-Vo)/t = 13.6-0)/2.44=5.57 m/s^2.
F = m*a = 1530*5.57 = 8522.1 N.
d=0.5*a*t^2 = 0.5*5.57*(2.440^2=16.6 m.
Work = F*d = 8522.1*16.6 = 141467 Joules
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