During a road test,a driver brakes a 1420kg car moving at 64.8km/h [W]. The car slows down and comes to a stop after moving 729m [W]. What is the net force acting on the car?
Well, this situation reminds me of a clown trying to stop a unicycle on a slippery banana peel! Quite the balancing act, I must say. Let's calculate the net force.
First, we need to convert the car's initial velocity from km/h to m/s. So, 64.8 km/h is approximately 18 m/s. Now, it's time to buckle up and apply some physics.
We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (18 m/s), a is the acceleration, and s is the distance covered.
Plugging in the values, we get 0 = (18 m/s)^2 + 2a(729 m). Solving for a, we find a ≈ -5.59 m/s^2.
Now, let's find the net force using Newton's second law: F = ma, where F is the net force, m is the mass (1420 kg), and a is the acceleration.
Thus, F = (1420 kg) * (-5.59 m/s^2), which gives us a net force of approximately -7924 N (since the force is in the opposite direction of motion, we get a negative value).
Therefore, the net force acting on the car is roughly -7924 N. Keep in mind, this is just an approximate value. You wouldn't want to rely on a clown's calculations for anything serious!
To find the net force acting on the car, we will use the equation of motion:
\[v_f^2 = v_i^2 + 2ad\]
Where:
- \(v_f\) is the final velocity (0 m/s since the car comes to a stop),
- \(v_i\) is the initial velocity (64.8 km/h),
- \(d\) is the distance covered (729 m), and
- \(a\) is the acceleration.
First, let's convert the initial velocity from km/h to m/s:
\[v_i = \frac{64.8 \times 1000}{3600} \, \text{m/s}\]
Next, rearranging the equation, we can solve for the acceleration:
\[a = \frac{v_f^2 - v_i^2}{2d}\]
Substituting the given values:
\[a = \frac{0 - \left(\frac{64.8 \times 1000}{3600}\right)^2}{2 \times 729}\]
Calculating this:
\[a \approx \frac{-104.969}{1458}\, \text{m/s}^2\]
Since the car is slowing down, the net force acting on the car is in the opposite direction (east).
Finally, we can find the net force by multiplying the mass of the car (\(m\)) by the acceleration (\(a\)):
\[F_{\text{net}} = m \cdot a\]
Substituting the given mass:
\[F_{\text{net}} = 1420 \times \frac{-104.969}{1458} \, \text{N}\]
Calculating this:
\[F_{\text{net}} \approx -101.976 \, \text{N}\]
Therefore, the net force acting on the car is approximately -101.976 N (east).
To find the net force acting on the car, we can use the equation:
Net force = mass * acceleration
First, let's convert the velocity of the car from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s.
So, the velocity of the car, v = 64.8 km/h * 0.2778 m/s per km/h = 17.9976 m/s
Next, we need to find the acceleration of the car. We can use the equation:
a = (vf - vi) / t
where vf is the final velocity, vi is the initial velocity, and t is the time taken.
As the car comes to a stop, the final velocity (vf) is 0 m/s. The initial velocity (vi) is 17.9976 m/s. The time taken (t) can be found using the given distance and velocity:
distance = (vi + vf) * t / 2
Using the given distance of 729 m and substituting the values, we can solve for t:
729m = (17.9976m/s + 0m/s) * t / 2
Simplifying the equation, we get:
1458 = 17.9976t
Dividing both sides by 17.9976, we find:
t ≈ 80.91 seconds
Now, we can calculate the acceleration (a):
a = (0m/s - 17.9976m/s) / 80.91s
Simplifying further, we get:
a ≈ -0.2224 m/s²
The negative sign indicates that the car is decelerating.
Finally, we can find the net force using the formula:
Net force = mass * acceleration
Substituting the given mass of the car (1420 kg) and the calculated acceleration, we get:
Net force = 1420 kg * (-0.2224 m/s²)
Calculating the expression, we find:
Net force ≈ -314.288 N
Therefore, the net force acting on the car is approximately -314.288 N, which means it is acting in the opposite direction to the motion of the car (westward).
vf^2=vi^2+2ad but a=forcenet/mass
change vi to m/s, solve for forcenet.