A milk carton with a mass of 2.00 kg is pulled across a table with a horizontal force of 3.00 N. If the coefficient of friction is 0.110, what is the acceleration of the milk carton? Show your work.

Question

A milk carton with a mass of 2.00 kg is pulled across  a table with a horizontal force of 3.00 N. If the coefficient of friction is 0.110, what is the acceleration of the milk carton? Show your work.

anonymous · Answer

@Loser66 any chance you could help with this?

anonymous · Answer

Firstly, you must calculate the Force produced by the friction which is given by:
$$F=\mu R$$.  mu is the coefficient of friction: 0.110 (dimensionless) and R is the 'normal reaction force' acting vertically upwards from the milk carton. The normal force is caused by the milk carton's weight. Newton's 3rd law tell us the weight has an equal but opposite force acting from the surface upwards, this is the normal force. 

Now to throw some numbers in to calculate the force due to friction:
$$F=(0.110). 19.62$$
Therefore force due to friction is 2.1582N.

This force works against the pulling force of 3.00N. Therefore, there is an overall force of 0.8418N acting in the direction you're pulling the carton

$$_{Fnet}=3.00 - 2.1582 = 0.8148 N$$

Now we know the net force acting on the carton, it's a simple case of using Newton's 2nd law to calculate the acceleration acting on a particle of 2Kg (19.62N =2g)

$$F=ma$$
$$0.8418=19.62.a$$
Therefore acceleration is:
$$a=0.0429 ms/s$$

It looks as if the question wants the answer to 3 sig figs too.
Hope this helps.

anonymous · Answer

@Dan.Harrison thank you so much!