A 1200-kg car going 30 m/s applies its brakes and skids to rest. If the friction force between the sliding tires and the pavement is 6000 N, how far does the car skid before coming to rest?
Can you help me get the formula to use .
Thanks
simon simon nigatu @ gamali.com
A 1200-kg car going 30 m/s applies its brakes and skids to rest. If the friction force between the sliding tires and the pavement is 6000 N, how far does the car skid before coming to rest?
To solve this problem, we can use the equations of motion. The equation we'll use is:
v^2 = u^2 + 2as,
where:
- v is the final velocity (0 m/s, because the car comes to rest),
- u is the initial velocity (30 m/s),
- a is the acceleration (which can be calculated using the friction force),
- s is the distance traveled.
The friction force between the tires and the pavement can be used to find the acceleration:
F = ma,
where:
- F is the friction force (6000 N),
- m is the mass of the car (1200 kg),
- a is the acceleration.
Rearranging the formula, we can solve for a:
a = F / m.
Substituting the given values, we have: a = 6000 N / 1200 kg = 5 m/s^2.
Now we can rearrange the first equation to solve for s:
s = (v^2 - u^2) / (2a).
Substituting the known values, we have:
s = (0^2 - 30^2) / (2 * 5).
Calculating this, we get: s = -900 / 10 = -90 m^2/s^2.
Since the distance cannot be negative, we take the magnitude of the answer: s = |-90| = 90 meters.
Therefore, the car will skid for a distance of 90 meters before coming to rest.
Set the work done against friction equal to the initial kinetic energy and solve for X
(1/2) M Vo^2 = 6000 * X
X = (1/1200)M Vo^2
Vo is the initial velocity, 30 m/s.