Calculating Initial Velocity: Physics of Home Alone 2's Tool Chest Scene

[Caronica]

Hi,

My friend and I are doing a physics project that analyzes movie physics and we have a quick question. In Home Alone 2, the kid ties a doorknob to a tool chest on wheels that is a floor up. The bad guys pull on the doorknob and the tool chest starts falling down the stairs. How would we calculate the initial velocity of the tool chest after it gets pulled by the rope?

 

[prasannapakkiam]

Welcome to Physics Forums.

Answer to your question: Is it not 0? The tool chest accelerates down he stairs.

 

[m2003]

Calculating the initial velocity of the tool chest in the Home Alone 2 scene can be done using the principles of Newton's laws of motion. We can assume that the tool chest is initially at rest before the rope is pulled, and that the only force acting on it is the tension force from the rope. This tension force will accelerate the tool chest downwards, causing it to gain velocity as it falls down the stairs.

To calculate the initial velocity, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, we know that the final velocity is 0 m/s since the tool chest comes to a stop at the bottom of the stairs. We also know that the acceleration due to gravity is 9.8 m/s^2. Therefore, we can rearrange the equation to solve for the initial velocity, u.

u = v - at

Plugging in the values, we get:

u = 0 m/s - (9.8 m/s^2)(t)

To find the time, t, we can use the equation s = ut + 1/2at^2, where s is the distance traveled. In this case, the distance traveled is the height of the stairs, which we can estimate to be around 3 meters. Therefore, we get:

3 m = (0 m/s)(t) + 1/2(9.8 m/s^2)(t^2)

Solving for t, we get t = 0.78 seconds.

Now, we can plug this value of t into our initial velocity equation to find the initial velocity of the tool chest:

u = 0 m/s - (9.8 m/s^2)(0.78 s)

u = -7.644 m/s

Therefore, the initial velocity of the tool chest after it is pulled by the rope in the Home Alone 2 scene is approximately 7.644 m/s downwards. Keep in mind that this is an ideal calculation and does not take into account any friction or air resistance. It also assumes that the rope is pulled with a constant force throughout the entire fall.