Find the coefficient of kinetic friction between a 4.3-kg block and the horizontal surface on...

Question:

Find the coefficient of kinetic friction between a 4.3-kg block and the horizontal surface on which it rests if an 80 Newton/meter spring must be stretched by 2.7 cm to pull the block with constant speed. Assume the spring pulls in a direction 13 degrees above the horizontal.

Friction :

Friction is the force that will cause the retardation of the object and the retardation is due to the fact that there is a contact force that will exert a force opposite to the applied force. The force of friction is dependent on the normal force and the coefficient of friction.

Answer and Explanation:

Given

Mass of the block is {eq}m=4.3\ kg {/eq}

Spring constant of the spring is {eq}k=80\ N/m {/eq}

Initial stretch to the spring is {eq}x=2.7\ cm {/eq}

The angle of inclination is {eq}\theta = 13^\circ {/eq}

Force applied by the spring is:

{eq}F=kx\\ F=80\times 0.027\\ F=2.16\ N {/eq}

Now for the horizontal component of force:

{eq}F_x=F\cos\theta\\ F_x=2.16\cos 13^\circ\\ F_x=2.104\ N {/eq}

Since the pull is for the constant speed hence the friction force will be equal to the applied force:

{eq}\mu (mg\cos\theta)=F_x\\ \mu (4.3\times 9.81\cos 13^\circ)=2.104\\ \mu=\frac{2.104}{41.1}\\ \mu=0.05 {/eq}

Thus, the coefficient of friction is 0.05


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