# AP Physics B 1.3 Newtonian Mechanics

AP® Physics B: Newtonian Mechanics Drill 1, Problem 3. With what acceleration does lunch arrive?

AP Physics B | Newtonian Mechanics |

AP Physics B/C | Newtonian Mechanics |

AP Physics C | Newtonian Mechanics |

Language | English Language |

### Transcript

to the upper basket. The heavier basket falls, and the lighter basket rises.

Lunchables are served! If we assume that the pulley is frictionless,

the basket of highly processed snack food has a mass of 2 kilograms, and the baskets

of rocks has a mass of 6 kilograms... ...with what acceleration does your lunch arrive?

And here are the possible answers...

Before we even do any calculations, we can eliminate every choice except for A or B.

Objects near the surface of the earth accelerate at 9.8 meters per second squared, which

can round to about 10.

However, since we have two baskets, accelerated

by gravity working against each other,

we know that the net acceleration MUST be less than 10.

We don't know exactly how much less, but we're left with A and B.

This is a pretty complex system, and finding

acceleration won't be as simple as eating that delicious lunch dear old Mom sent you.

But a little complexity never stopped us... To find acceleration, we need to look at each

basket and the forces acting on that basket.

To start, we'll look at the one with the food in it.

The basket has two forces on it: Gravity and tension.

Gravity points down, as always, and tension points up.

Now, when we're solving a physics problem, we can't go wrong with F = m times a.

We know that the net force on this object is equal to its mass times its acceleration,

which we're trying to find. The net force is also equal to the sum of

the forces acting on the basket.

Tension and gravity are working in opposite directions,

so one must be negative.

Since our lunch is accelerating up... which hopefully won't happen after we eat it...

...we will call that one positive, and down will be negative.

Our net force is then also equal to T minus m times g.

The second basket, the one with the rocks,

is in a similar situation. We use F = m times a to find that the net

force on the basket with the rocks is also equal to its mass times acceleration, and

the sum of the forces acting on it. But wait! We almost made a terrible mistake.

For the basket with the lunch, we made acceleration upwards positive, so we have to keep it consistent...

The basket with the rocks is accelerating downwards, so we have to make acceleration

negative. Our equation then looks like this: We now have two equations that we know to

be true about this system: We know the mass of the two baskets, and the

force of gravity... which leaves only acceleration, which we're trying to find... and tension.

We know that acceleration and tension must be the same for the two baskets because they're

part of the same system.

This leaves us with a system of two equations for us to solve.

To begin, we isolate T for both equations.

We add m times g to both sides of both equations, and we get the following:

Then, we set the two equations equal to each other, so we get that the mass of lunch times

acceleration... plus the mass of lunch times the acceleration due to gravity... is equal

to the negative mass of the rocks times acceleration... plus the mass of the rocks times gravity.

Plugging in values, we get 2a plus 2 times 10 is equal to negative 6 times a plus 6 times 10.

We add 6a to both sides and subtract 20 from both sides to get 8a is equal to 40.

Finally, we divide both sides by 8 to get

the acceleration is equal to 5 meters per second squared.

So the basket accelerates at 5 meters per second squared...

...which is answer B. As in, "Barely edible."