Stopping distance given frictional force

[Chica1975]

Homework Statement

In an emergency stop on a level, dry concrete road, the magnitude of the friction force when sliding is approx 80% of the weight of the vehicle what is the stopping distance required for a vehicle traveling at 88km/h (24.444m/s). assume that all the wheels lock when the brakes are applied.

Homework Equations

to be honest i freaked out when I saw this question - i looked at my textbook but it really is useless.
is the frictional coefficient .20?

The Attempt at a Solution

I have no idea what to do
apparently the answer is 38m

 

[berkeman]

Homework Statement

In an emergency stop on a level, dry concrete road, the magnitude of the friction force when sliding is approx 80% of the weight of the vehicle what is the stopping distance required for a vehicle traveling at 88km/h (24.444m/s). assume that all the wheels lock when the brakes are applied.

Homework Equations

to be honest i freaked out when I saw this question - i looked at my textbook but it really is useless.
is the frictional coefficient .20?

The Attempt at a Solution

I have no idea what to do
apparently the answer is 38m

Not 0.2. Remember that the frictional force is F = mu * N, where in this case, N is the weight of the car. So what is mu?

Anyway, if the frictional force is constant (which it is in this problem), you use the kinematic equations of motion for constant accerleration (just like you use for gravity-related problems). What is the equation that relates the velocity to the applied acceleration (or deceleration in this case)?

 

[Chica1975]

I have no idea what mu is and N I don't have just that vehicle is traveling at 88km/h. I am ok with the kinematic equations when I have the numbers to plug in. I have no idea how to get the N for the weight of the vehicle just by being given that the frictional force is 80% the weight of the vehicle.

 

[Chica1975]

can someone please tell me how to do this? I have no idea how to workout the weight of the car. I need to hand this in. Without it being explained I really don't know what i am doing in the least.

 

[rwisz]

I would approach this problem by first identifying the known variables and equations that can help solve it. The known variables in this case are the speed of the vehicle (88km/h) and the friction force (80% of the weight of the vehicle). The unknown variable is the stopping distance.

To solve for the stopping distance, we can use the equation for kinetic energy: KE = 1/2mv^2, where m is the mass of the vehicle and v is the initial velocity. We can also use the equation for frictional force: Ff = μN, where μ is the coefficient of friction and N is the normal force, which is equal to the weight of the vehicle.

Using these equations, we can set up the following equation:

KE = Ff * d

Where d is the stopping distance. We can rearrange this equation to solve for d:

d = KE / Ff

Substituting the known values, we get:

d = (1/2mv^2) / (μN)

Since we are given that the friction force is 80% of the weight of the vehicle, we can substitute 0.8N for Ff, giving us:

d = (1/2mv^2) / (0.8N)

We also know that the mass of the vehicle does not change, so we can simplify further:

d = (1/2m) * (v^2 / 0.8N)

To solve for d, we need to find the value for N, which is the weight of the vehicle. We can use the formula F = ma (Newton's second law) to find the weight:

F = ma

N = mg

Where g is the acceleration due to gravity, which is approximately 9.8 m/s^2. So, we can rewrite the equation as:

d = (1/2m) * (v^2 / 0.8mg)

Plugging in the known values, we get:

d = (1/2 * 0.8 * 24.444^2) / (0.8 * 9.8)

Solving for d, we get a stopping distance of approximately 38 meters. This means that in an emergency stop on a level, dry concrete road, a vehicle traveling at 88km/h will require approximately 38 meters to come to a complete stop, assuming all wheels are locked