# An object with a mass of #4 kg# is lying still on a surface and is compressing a horizontal spring by #1/4 m#. If the spring's constant is #6 (kg)/s^2#, what is the minimum value of the surface's coefficient of static friction?

##### 1 Answer

#### Answer:

#### Explanation:

We will solve this first by setting up a force diagram and writing out sum of forces statements.

Where

#vecn# is the normal force,#vecF_g# is the force of gravity,#vecF_(sp)# is the force of the spring, and#vecf_s# is the force of static friction. I will define rightward as positive.

Statements of the net force:

#sumF_x=vecF_(sp)-vecf_s=mveca_x#

#sumF_y=vecn-vecF_g=mveca_y#

The object is not moving, and experiences no net acceleration in either

#sumF_x=vecF_(sp)-vecf_s=cancel(mveca_x)=vec0#

#sumF_y=vecn-vecF_g=cancel(mveca_y)=vec0#

We can rearrange these statements to find:

#vecF_(sp)=vecf_s#

#vecn=vecF_g# Where

#vecF_g=mg#

The equation for the frictional force is

#vecF_(sp)=mumg#

The equation for the spring force is given by Hooke's law as

#k(Deltas)=mumg#

Solving for

#mu=(k(Deltas))/(mg)#

Using our known values, we can calculuate

#mu=((6N/m)(1/4m))/(4kg*9.8m/s^2)#

#mu=0.038...~~0.04#

Therefore, the minimum coefficient of static friction to keep the object still is