Question
A 1260-kg car starts from rest at the bottom of a drive way and has a speed of 3.00 m/s at a point where the drive way has risen a vertical height of 0.600 m. Friction and the drive force produced by the engine are the only two nonconservative forces present. Friction does -2870 J of work. How much work does the engine do?
Answers
Elena
W=PE+KE+W(fr) =mgh+mv²/2 +W(fr) =
=m(gh+ v²/2)+W(fr) =1260(0.6•9.8 +3²/2)+2870 =
=13078.8+2870 =15948.8 J
=m(gh+ v²/2)+W(fr) =1260(0.6•9.8 +3²/2)+2870 =
=13078.8+2870 =15948.8 J