Question

this is a 2 part question.18) A backpack full of books weighing 52.0 N rests on a table in a physics laboratory classroom. A spring with a force constant of 150 N>m is attached to the backpack and pulled horizontally, as indicated in FIGURE 6-37. (a) If the spring is pulled until it stretches 2.00 cm and the pack remains at rest, what is the force of friction exerted on the backpack by the table? (b) Does your answer to part (a) change if the mass of the backpack is doubled? Explain.

Answer 1

(a) 3 N

(b) No, it does not.

Step-by-step explanation

Parameters given:

Weight of backpack, W = 52N

Spring constant of speing, k = 150 N/m

Extension of spring, x = 2.00 cm = 0.02 m

(a) We want to find the force of friction exerted on the backpack by the table.

Let us consider the forces acting in the horizontal direction. Since the backpack did not move, it means that there is no acceleration by the backpack, hence, the net horizontal force is 0:

`F_x=ma=0`

There are two forces acting on the backpack when the spring is pulled namely:

-> Force of the pull of the spring, F

-> Force of static friction, Fs

Taking the right hand to be the positive direction, the friction force acts in the positive direction while the pull acts in the negative direction.

Therefore, we have that:

`F_s-F=ma=0`

The force of the pull is given by Hooke's law:

`F=kx`

Therefore, we have that:

`\begin{gathered} F_s-F=0 \\ F_s-kx=0 \\ \Rightarrow F_s=kx \\ \Rightarrow F_s=150\cdot0.02 \\ F_s=3N \end{gathered}`

That is the frictional force exerted on the backpack by the table.

(b) In the calculation above, we see that the friction force is not dependent on the mass of the backpack. This implies that if the mass of the backpack is doubled, the friction force does not change.

Hence, the answer is no.