how much force does a bear weighing 300 kilograms and moving at 30 k/s generate

Answer 1

Hello Skye, let's first delve into the concept of force and stopping distance using a few analogies and examples.

1. Short Stopping Distance:

Imagine a car moving at a high speed on a highway and suddenly needing to come to a stop. If the driver applies the brakes and the car comes to a stop relatively quickly, the force experienced by the occupants is significant. This is due to a high deceleration, and the force exerted by the brakes on the car is what brings it to a halt.

In the case of the bear, if it were to stop within a short distance (let's say 1 meter), the force required to bring it to a stop is substantial. It's similar to a quick stop of a car – the deceleration is high, leading to a more noticeable force.

2. Long Stopping Distance:

Now, let's consider the scenario where the car has more space to gradually come to a stop. The driver applies the brakes, but this time the car slows down over a much longer distance, perhaps 100 meters. In this case, the force experienced by the occupants is less noticeable because the deceleration is lower. The brakes still exert a force to decelerate the car, but it's spread out over a longer distance.

Applying this idea to the bear, if it were to stop over a longer distance, the force required would be less noticeable. The gradual deceleration reduces the force experienced, making it more akin to a gentle stop.

In Summary:

1. Short Stopping Distance: Higher deceleration, more noticeable force.
2. Long Stopping Distance: Lower deceleration, less noticeable force.

Mathematical model

Now let's delve into the equations and put numbers into our words:

1. The force is calculated using Newton's second law of motion:

(1) F=m⋅a

Assuming the bear comes to a stop, and let's say it does so over a distance of s meter, you can use the following kinematic equation to find the acceleration:

(2) Velocity equation: v = a⋅t + u

(3) Position equation: s = 1/2⋅a⋅t² + u⋅t

by solving (2) and (3) we get: a = - u²/(2s)

where:

1. v is the final velocity (0 m/s, as the bear comes to a stop),
2. u is the initial velocity (30 m/s),
3. a is the acceleration,
4. s is the distance (final position minus initial position)

Plugging into (1) we get: F = - m⋅u²/(2s)

Where the negative sign indicates that the force is against the direction of motion (a stopping force).

I'm also guessing, you had a typo in the speed and it's 30 km/h = 8.33 m/s. (For reference, Voyager 1 is cruising at 17km/s, and earth is cruising around the sun at 30km/s)

Let's look into the 2 cases:

1. Short stopping distance (1m): F = 10.4 kN (or around 1ton of force).
2. long stopping distance (10m): F = 1.04 kN (or around 100kg of force - equivalent to a strong punch in the face).

Now, if you really meant 30km/s and assuming a short stopping distance of 1m then F = 135,000,000,000 N (or 13.5M tons).