Question
A 1.92 × 103 kg car accelerates uniformly from rest to 12.4 m/s in 3.77 s.
What is the work done on the car in this time interval?
What is the work done on the car in this time interval?
Answers
a=(V-Vo)/t = (12.4-0)/3.77 = 3.29 m/s^2
d = 0.5*3.29*3.77^2 = 23.37 m.
Work = F*d = (m*g) * d=(1920*9.8)23.37 =
439,805 Joules.
d = 0.5*3.29*3.77^2 = 23.37 m.
Work = F*d = (m*g) * d=(1920*9.8)23.37 =
439,805 Joules.
Related Questions
1. A car starts from rest and accelerates uniformly at 3.0 m/s2. A second car starts from rest 6.0 s...
A particle P, starting from rest at A, moves in a straight line ABCD. It accelerates uniformly at 5m...
A particle accelerates uniformly from rest at 6.0ms for 8s and then decelerate uniformly to rest in...
A particle accelerates uniformly from rest at 6.oms for 8s and then decelerates uniformly to rest in...