Unit 2 motion problem

[Stick]

A 0.50kg metal block hanging off the side of a desk is attached by a piece of string to a dynamics cart. The block is allowed to fall to the ground from rest, dragging the cart along. The mass of the cart is 2.5kg.

a) If friction is ignored, what is the acceleration of the block as it falls?
b) How fast will be block be travelling after 0.50s?
c) If a frictional force of 4.3N acts on the cart, what is its acceleration?

It would be great if you could show me how to solve these problems. Thanks for your help.

[Lasercookie]

The first thing you should do is talk yourself through the scenario here and picture it in your head. Drawing a picture is always good too.

a)  So the block experiences a force such that (as it is the one that drops) 
It's acceleration just isn't g though, it's got the cart dragging behind it, so we can't forget that the force has to accelerate that total mass.

So

b) const. accel:

c) So let's picture the scenario. The block and the cart fall, and it'll experience a force such that
We have a frictional force of 4.3N acting upon it, so the force it'll experience will be given by

[Bhootnike]

Some working out i found if you want generic type answer:

> "a) If friction is ignored, what is the acceleration of the block as it falls?"

The acceleration equal the net force divided by the mass (that's Newton's 2nd Law).

a = Fnet / m_block

Fnet is a combination of the downward force of gravity (m_block)(g) and the the upward force of the string's tension (T)

a = ((m_block)(g)-T)/m_block

But you don't know how much "T" is. That's where the cart comes in. The CART'S acceleration is:

a_cart = (Fnet on cart) / m_cart

But the only force on the cart is the string's tension "T":

a_cart = T / m_cart

AND, the block and the cart are both accelerating at the same rate (because they're tied together); so use the same "a" variable for both:

a = T / m_cart

Recap:
------------
a = ((m_block)(g)-T)/m_block
a = T / m_cart

Those are 2 equations in 2 unknowns ("a" and "T"). Use the algebra of simultaneous equations to solve.

> "b) How fast will the block be travelling after .5s"

V = at = a(.5s) [Use the "a" you calculated above]

> "c) If a frictional force of 4.3N acts on the cart, what is its acceleration?"
This means you'd need to change this equation:
a = T / m_cart

to this:
a = (T - 4.3N) / m_cart

The rest is the same; solve the simultaneous equations.

[Stick]

Thanks for your help, laseredd and Bhootnike.