No.
Your equations are troubling, as it reveals you have no idea what you are doing.
Work input= KE output
Netforce*distance= 1/2 m v^2
4500*100= 1/2 *1400*v^2
Solve for v. This is about ten times what you thought.
W=FnetX distance=ma x distance
4500N=(1/2)1400kg V^2
2.5m/s=V
is this right? if not what am i doing wrong?
Your equations are troubling, as it reveals you have no idea what you are doing.
Work input= KE output
Netforce*distance= 1/2 m v^2
4500*100= 1/2 *1400*v^2
Solve for v. This is about ten times what you thought.
First, let's calculate the acceleration (a) of the car using Newton's second law:
F_net = ma
4500N = 1400kg * a
a = 4500N / 1400kg
a ≈ 3.214 m/s²
Now, we can use the kinematic equation to find the final velocity (V) after the car has traveled a distance of 100m from rest:
V^2 = u^2 + 2as
Where:
u = initial velocity (0 m/s as the car starts from rest)
s = distance (100m)
a = acceleration (3.214 m/s²)
Plugging in these values:
V^2 = 0^2 + 2 * 3.214 m/s² * 100m
V^2 = 0 + 642.8 m²/s²
V ≈ √(642.8) m/s
V ≈ 25.36 m/s
So, the speed of the car after traveling 100m is approximately 25.36 m/s.
To find the final speed (V) of the car after traveling 100m, we can use the equations of motion.
Step 1: Identify the known quantities:
- Mass of the car (m) = 1400 kg
- Net force acting on the car (Fnet) = 4500 N
- Distance traveled by the car (d) = 100 m
- Initial velocity of the car (u) = 0 m/s (since the car starts from rest)
Step 2: Calculate the acceleration (a) using Newton's second law: Fnet = ma
a = Fnet / m
a = 4500 N / 1400 kg
a ≈ 3.21 m/s^2
Step 3: Use the equation of motion to calculate the final velocity (V):
V^2 = u^2 + 2ad
Substituting the values,
V^2 = (0 m/s)^2 + 2 × 3.21 m/s^2 × 100 m
V^2 = 2 × 3.21 m/s^2 × 100 m
V^2 = 642 m^2/s^2
Taking the square root of both sides to find V:
V ≈ √(642 m^2/s^2)
V ≈ 25.34 m/s
So, the correct final speed of the car after traveling 100m is approximately 25.34 m/s.