Question

An applied pulling force of 100 N is used to accelerate an object to the right along a rough surface that offers 40 N of frictional resistance if the normal force is 60N what is the acceleration of the object

Answer 1

Answer:

$9.8 m/s^2$

Explanation:

First of all, we analyze the vertical direction.

Along this direction, the object is in equilibrium, so its acceleration is zero, and therefore  the net force on it is zero. There are only two forces acting on the object vertically: the weight, W (downward), and the normal force, N (upward), and since the net force must be zero, we can write

$W=N$

And since N = 60 N, the weight of the object is

W = 60 N

From this, we can also find the mass of the object, using the equation:

$W=mg \rightarrow m = \frac{W}{g}=\frac{60}{9.8}=6.1 kg$

where $g=9.8 m/s^2$ is the acceleration of gravity.

Now we analyze the forces along the horizontal direction. We have:

A pulling force of F=100 N forward

A frictional resistance of $F_f = 40 N$ backward

So the equation of motion in this direction is

$F-F_f = ma$

And solving for a, we find the acceleration of the object:

$a=\frac{F-F_f}{m}=\frac{100-40}{6.1}=9.8 m/s^2$

in the horizontal direction, forward.