Problem 28 - College Physics

Question

A cat drops from a shelf \(4.0 \mathrm{ft}\) above the floor and lands on all four feet. His legs bring him to a stop in a distance of \(12 \mathrm{~cm}\). Calculate (a) his speed when he first touches the floor (ignore air resistance), (b) how long it takes him to stop, and (c) his acceleration (assumed constant) while he is stopping, in \(\mathrm{m} / \mathrm{s}^{2}\) and \(g\) 's.

Step-by-Step Solution

Answer

(a) 4.89 m/s, (b) 0.049 s, (c) 99.78 m/s² or 10.17 g.

Step 1: Convert initial height to meters

First, convert the height from feet to meters because we'll need it in SI units for further calculations. Given height: \(4.0\) feet.We know that: \(1\) foot = \(0.3048\) meters.Therefore, \(4\) feet = \(4 \times 0.3048 = 1.2192\) meters.

Step 2: Calculate the cat's speed when it first touches the floor

Use the formula for the final velocity of a falling object:\[ v = \sqrt{2gh} \]where :- \( g \) is the acceleration due to gravity, \(9.81 \ \text{m/s}^2\), and- \( h \) is the height of \(1.2192\) meters.Substituting the values:\[v = \sqrt{2 \times 9.81 \times 1.2192} \approx 4.89 \ \text{m/s}\]Thus, the speed when the cat first touches the floor is approximately \(4.89 \ \text{m/s}\).

Step 3: Convert stopping distance to meters

The problem states the cat stops over a distance of \(12\, \text{cm}\). First, convert this distance into meters:\( 12 \) cm = \( 0.12 \) meters.

Step 4: Calculate the time it takes the cat to stop

The formula to use is derived from the final velocity equation of motion:\[ v = u + at \]where :- \( v \) is the final velocity (\(0 \ \text{m/s}\) when the cat stops),- \( u \) is the initial velocity which is the speed when the cat first touches the floor, \(4.89 \ \text{m/s}\), and- \( a \) is the acceleration.Rearranging:\[ t = \frac{v - u}{a} \]Acceleration can be found using the equation:\[ v^2 = u^2 + 2as \]Solving for \( a \):\[ a = \frac{(v^2 - u^2)}{2s} \]Substituting the known values:\[ a = \frac{(0^2 - 4.89^2)}{2 \times 0.12} \approx -99.78 \ \text{m/s}^2 \]The negative sign indicates deceleration. Now use this to find the time:\[ t = \frac{0 - 4.89}{-99.78} \approx 0.049 \ \text{seconds}\]

Step 5: Calculate the magnitude of acceleration in terms of g

To find acceleration in terms of \( g \), divide the magnitude of acceleration by \( g \):\[ a_{g} = \frac{99.78}{9.81} \approx 10.17 \ g \]

Step 6: Conclusion

The speed when the cat first touches the floor is approximately \(4.89 \ \text{m/s}\). It takes \(0.049\) seconds for the cat to stop. The acceleration while stopping is \(99.78\ \text{m/s}^2\) or approximately \(10.17 \ g\).

Key Concepts

Physics problem-solving

When approaching a physics problem, it’s crucial to break it down into manageable steps. This not only simplifies complex concepts but also ensures no detail is overlooked. Let’s look at a typical physics problem involving free fall motion. You should start by identifying the given data and what you need to find. With our cat example, the heights and distances are given, enabling us to compute its velocity, acceleration, and time it takes to stop.

• Convert all measurements to the same unit system, usually SI units (meters, seconds).
• Identify relevant formulas that tie into your known values and what you need to calculate.
• Perform calculations step-by-step, recording your values with unit verification at each stage.
• Review results for consistency and feasibility compared to your expectations.

This problem-solving approach applies to a vast array of physics topics, helping clarify and systematically tackle any problem that comes your way.

Free fall motion

Free fall motion refers to the motion of objects falling under the influence of gravity alone, with no other forces acting upon them, such as air resistance. In our problem, the cat dropping from a shelf is a good example. All objects in free fall near the surface of the Earth accelerate downwards at a constant rate, typically denoted as \( g = 9.81 \, \text{m/s}^2 \).

One of the key equations for free fall motion, which we used, is \( v = \sqrt{2gh} \), where \( v \) is the final velocity at which the object hits the ground, and \( h \) is the height from which it falls. Knowing the constant acceleration \( g \), you can predict the falling object's complete behavior, from the speed it reaches to how long it takes to fall.

Remember, in the absence of air resistance, all objects, irrespective of their mass, fall at the same rate. This fundamental principle is essential not only for this problem but also forms the basis for understanding more complex dynamics of objects in motion.

Acceleration

Acceleration is the rate of change of velocity of an object and can indicate speeding up, slowing down, or changing direction. In our problem, we focus on two types: the acceleration due to gravity (9.81 \( \text{m/s}^2 \)) during free fall, and the cat's deceleration as it hits the ground and comes to a stop.

We used the equation for acceleration when stopping, \( a = \frac{(v^2 - u^2)}{2s} \), where \( v \) is the final velocity, \( u \) is the initial velocity right before touching the ground, and \( s \) is the stopping distance. For the cat, the calculated acceleration was \(-99.78 \text{m/s}^2\).

In physics, acceleration is often expressed as multiples of \( g \), the gravitational constant. This provides a relatable measure of acceleration, especially in contexts such as human tolerance to acceleration forces. Our example showed an acceleration of approximately \( 10.17 \, g \), indicating a significant deceleration force.

Unit conversion

Unit conversion is an essential skill in physics problem-solving. It ensures that all measurements are in compatible units throughout calculations, reducing the risk of errors. In our scenario, it was crucial to convert feet to meters and centimeters to meters before proceeding with the calculations.

The conversion process is straightforward:

• Convert feet to meters: 1 foot equals 0.3048 meters. Therefore, multiply the number of feet by 0.3048 to get meters.
• Convert centimeters to meters: 1 centimeter equals 0.01 meters. Thus, multiply the number of centimeters by 0.01 to get meters.

Converting all measurements to SI units allows the use of standard formulas and constants, such as the acceleration due to gravity \( g \) in meters per second squared. This not only aids consistency and accuracy but also facilitates clearer communication of scientific information, as SI units are universally recognized.