m =1280 kg, v =80 km/h = 22.2 m/s, t = 23 s.
W(net) = ΔKE =KE(final) – KE(initial) =
= m•v²/2 -0=1280•(22.2)²/2 =3.154•10^5 J.
v = a•t , a =v/t = 22.2/23 = 0.97 m/s²,
s= a•t²/2 = 0.97•23²/2 = 257 m.
W(net) = F(net) •s,
F(net) = W(net)/s =3.154•10^5/257 = 1227 N.
F(net) = F(applied) – F(fr) ,
F(applied) = F(net) + F(fr) = 1227+449 = 1676 N.
A 1280-kg car starts from rest and accelerates to 80 km/h in 23.0 s. Friction exerts an average force of 449 N on the car during this time.
(a) What is the net work done on the car?
J
(b) How far does the car move during its acceleration?
m
(c) What is the net force exerted on the car during this time?
N
(d) What is the forward force exerted on the car as a result of the engine, power train, and wheels pushing backward on the road?
2 answers
these are all wrong :(
3 answers