An automobile weighing 15000n crashes into a wall with a velocity of 5m/sec. If the car moves .15m before coming to a stop, what average force does the car exert? on the wall?
Force*timeofimpact=mass(change of velocity)
To find the average force exerted by the car on the wall, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, we need to find the acceleration first.
We can use the equation of motion: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Given:
Mass of the car (m) = 15000 N
Initial velocity (u) = 5 m/s
Distance traveled (s) = 0.15 m
First, let's find the final velocity (v) using the equation of motion:
v^2 = u^2 + 2as
v^2 = (5 m/s)^2 + 2 * a * 0.15 m
v^2 = 25 m^2/s^2 + 0.3 a
v^2 = 0.3 a + 25 m^2/s^2
Since the car comes to a stop, the final velocity (v) is equal to 0 m/s:
0 = 0.3 a + 25 m^2/s^2
0.3 a = -25 m^2/s^2
Now, let's solve for the acceleration (a):
a = (-25 m^2/s^2) / 0.3
a ≈ -83.33 m/s^2
Since the acceleration is negative, it means the car is decelerating.
Now we can find the average force exerted by the car on the wall using Newton's second law of motion:
F = m * a
F = 15000 N * (-83.33 m/s^2)
F ≈ -1,250,000 N
The negative sign indicates that the force exerted by the car on the wall is in the opposite direction of the motion.
Therefore, the average force exerted by the car on the wall is approximately -1,250,000 Newtons.