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?
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.
these are all wrong :(
To find the answers to these questions, we need to use some basic physics equations.
(a) To calculate the net work done on the car, we can use the work-energy principle, which states that the net work done on an object is equal to its change in kinetic energy. The formula is:
Work = (1/2) * m * v^2
where m is the mass of the car and v is its final velocity.
In this case, m = 1280 kg and v = 80 km/h = 22.2 m/s. Plugging these values into the formula, we get:
Work = (1/2) * 1280 kg * (22.2 m/s)^2 = 317,619.2 Joules
Therefore, the net work done on the car is approximately 317,619.2 Joules (J).
(b) To find the distance the car moves during its acceleration, we can use the formula for average acceleration:
Average acceleration = (Change in velocity) / (Time taken)
In this case, the car starts from rest, so its initial velocity is 0 m/s. The final velocity is 22.2 m/s, and the time taken is 23.0 seconds. Plugging these values into the formula, we get:
Average acceleration = (22.2 m/s - 0 m/s) / 23.0 s = 0.965 m/s^2
Now, we can use the following kinematic equation to find the distance (s):
s = (1/2) * a * t^2
where a is the average acceleration and t is the time taken.
Plugging in the values, we get:
s = (1/2) * 0.965 m/s^2 * (23.0 s)^2 = 249.52 meters (m)
Therefore, the car moves approximately 249.52 meters during its acceleration.
(c) To find the net force exerted on the car during this time, we can use Newton's second law of motion:
Net force = mass * acceleration
In this case, the mass of the car is 1280 kg and the average acceleration is 0.965 m/s^2. Plugging these values into the formula, we get:
Net force = 1280 kg * 0.965 m/s^2 = 1235.2 Newtons (N)
Therefore, the net force exerted on the car during this time is approximately 1235.2 Newtons.
(d) The forward force exerted on the car as a result of the engine, power train, and wheels pushing backward on the road can be found by subtracting the force of friction from the net force.
In this case, the force of friction is given as 449 N. So the forward force can be calculated as:
Forward force = Net force - Force of friction
Forward force = 1235.2 N - 449 N = 786.2 Newtons (N)
Therefore, the forward force exerted on the car as a result of the engine, power train, and wheels pushing backward on the road is approximately 786.2 Newtons.