Question

Students work together during an experiment about Newton’s laws. The students use a setup that consists of a cart of known mass connected to one end of a string that is looped over a pulley of negligible friction, with its other end connected to a hanging mass. The cart is initially at rest on a horizontal surface. Students have access to common laboratory equipment to make measurements of components of the system. The students double the mass that hangs from the string. They also replace the original cart with a new cart that has double the mass. By doubling both masses, how will the tension in the string and the acceleration of the cart change?

Answer 1

Answer:

The tension will double and the acceleration will remain the same.

Explanation:

In order to solve this problem we must start by doing a drawing of the situation and by drawing free bodies diagrams for both elements. (See attached picture)

So let's analyze the first free body diagram. We can suppose the friction between the cart and the horizontal surface is zero and since we only care about the horizontal movement, we only take the horizontal forces into account. So we do a sum of forces:

$\sum F_{1}=m_{1}a$

since the only horizontal force for the first mass is the tension, we can say that:

$T=m_{1}a$

Now we can analyze the second mass. We will suppose the positive direction of the movement will be in the downwards direction, so we do a sum of forces:

$\sum F_{2}=m_{2}a$

The second mass will be affected by two forces, which are the force of gravity and the tension, so the sum of forces will be:

$-T+m_{2}g=m_{2}a$

Now we can combine both equations, so we get:

$-m_{1}a+m_{2}g=m_{2}a$

and now we can solve for the acceleration:

$m_{1}a+m_{2}a=m_{2}g$

$a(m_{1}+m_{2})=m_{2}g$

$a=\frac{m_{2}g}{m_{1}+m_{2}}$

This ratio will represent the acceleration with the original masses. But, what happens if the masses double? Let's find out:

$a_{double masses}=\frac{2m_{2}g}{2m_{1}+2m_{2}}$

we can factor a 2 from the denominator of the fraction so we get:

$a_{double masses}=\frac{2m_{2}g}{2(m_{1}+m_{2})}$

we can simplify it so we get:

$a_{double masses}=\frac{m_{2}g}{m_{1}+m_{2}}$

as you may see this is the same as the original acceleration we had found, so the acceleration remains the same.

What about the tension? We take the equation from the first sum of forces and double the mass.

$T_{double}=2m_{1}a$

since the original tension was:

$T=m_{1}a$

this means that when doubling the first masses, then the tension will also be doubled.