# Why does the heavier car stop sooner?

• scooby07
In summary, the question of which car will stop sooner on a slippery road depends on the combination of factors such as weight, friction, and momentum. The heavier car may have increased weight and momentum, but friction is independent of weight and can retard the speed of both cars equally. The deciding factor of who will travel the furthest is the ratio of kinetic energy to mass, with both cars stopping at the same time if they are identical. However, if one car is heavier, it may use more gas to attain the same speed as the lighter car. The important point to note is that sliding friction depends linearly on weight, but the deceleration of the car does not depend on mass.
scooby07

## Homework Statement

Given two cars, one heavier than the other but identical in all other respects, traveling at the same speed on a snowy or icy road which one will stop sooner?

Momentum would say that the lighter car should stop sooner but how does the heavier car's increased weight in combination with friction figure into the stopping distance? Shouldn't the heavier car be able to dig into snow more increasing the opposing frictional force? On ice even though neither car can "dig in", wouldn't the heavier car experience a greater frictional force causing it to stop sooner or is the increased momentum greater than this increase friction?

Explanations are fine as are equations. Thanks for the help.

One dependency is how much the angular momentum at the wheels is shifted into linear velocity. . .

p.s. this is a true driver dependency.

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How about I simplify the problem.

Let's just say neither car has ABS and to stop both drivers choose to mash on the brakes and lock up the wheels.

So essentially both cars are just sliding blocks.

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Frictional force is independent of weight, but dependent on material composition.

Momemtum is mass * velocity. p = mv. Friction will retard speed.

Friction uses energy directly. Kinetic energy(KE) is (m * v * v)/2.
So let us say KE = mK for this example. Friction will leave r*m*k units of energy to continue the work.

With one mass m1 and the other m2, the KE ratio, yields m1/m2 as the deciding factor of who travels the furthest. Example, if both cars are identical, then m1 = m2, and they both stop at the same time. This makes sense because you know heavier cars use more gas to attain the same speeds as lighter cars and since friction is independent of weight, the solution is complete.

The important point is at the beginning of this post.

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basePARTICLE said:
Frictional force is independent of weight, but dependent on material composition.

Momemtum is mass * velocity. p = mv. Friction will retard speed.

Friction uses energy directly. Kinetic energy(KE) is (m * v * v)/2.
So let us say KE = mK for this example. Friction will leave r*m*k units of energy to continue the work.

With one mass m1 and the other m2, the KE ratio, yields m1/m2 as the deciding factor of who travels the furthest. Example, if both cars are identical, then m1 = m2, and they both stop at the same time. This makes sense because you know heavier cars use more gas to attain the same speeds as lighter cars and since friction is independent of weight, the solution is complete.

The important point is at the beginning of this post.

This doesn't make sense. Sliding friction depends linearly on weight. F = k.mg
where k is the coefficient of friction. If you combine that with F=m.a to compute the deceleration of the car, you'll see that THAT doesn't depend on mass.

## 1. How does friction stop a car?

The brakes on a car use friction to slow down and eventually stop the car. When you press the brake pedal, it applies pressure to the brake pads which then press against the rotors. This creates friction and helps to slow down the car.

## 2. What factors affect the amount of friction needed to stop a car?

The weight of the car, the speed at which it is traveling, and the condition of the brakes and tires all play a role in the amount of friction needed to stop a car. The heavier the car and the faster it is going, the more friction is needed to bring it to a stop.

## 3. Why is it important to maintain proper tire pressure for effective friction?

Proper tire pressure ensures that the tires have enough contact with the road surface to create the necessary friction for stopping a car. If the tires are underinflated, there is less surface area in contact with the road, resulting in less friction and potentially longer stopping distances.

## 4. Can friction ever be a bad thing when it comes to stopping a car?

In certain situations, such as driving on wet or icy roads, too much friction can actually be a bad thing. This can lead to skidding and loss of control of the car. In these cases, anti-lock braking systems (ABS) are designed to prevent excessive friction and maintain control of the vehicle.

## 5. How can friction be minimized to improve fuel efficiency?

Reducing friction between the moving parts of a car, such as the engine and transmission, can improve fuel efficiency. This is achieved through the use of lubricants, such as oil, which reduce the amount of friction and heat generated by these parts. Low rolling resistance tires can also reduce friction and improve fuel efficiency.

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