An object with a mass of #6 kg# is on a plane with an incline of # - pi/6 #. If it takes #18 N# to start pushing the object down the plane and #1 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

#mu_s =( 3sqrt3)/10#
# mu_k = 1/(20sqrt6)#

#f = mu N#
#N = mgcos theta = mg sqrt3/2#

Now that we have our prerequisites, let's start by thinking about the static case.

#18 = mu mgsqrt3/2# #18 = mu 4 * 10 sqrt3/2#

When we solve, we

#mu_s =( 3sqrt3)/10#

Examining the kinetic scenario first

#1 = mu_k * 4*10*sqrt 3/2#

When we solve, we obtain

# mu_k = 1/(20sqrt6)#
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Answer 2

The coefficient of static friction is μs = 0.577 and the coefficient of kinetic friction is μk = 0.100.

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Answer 3

The force required to start pushing the object down the plane is the force of static friction, ( F_s ). The force required to keep pushing it once it's in motion is the force of kinetic friction, ( F_k ).

The force of static friction is given by:

[ F_s = \mu_s \cdot N ]

where ( \mu_s ) is the coefficient of static friction and ( N ) is the normal force.

The normal force ( N ) is equal to the gravitational force acting perpendicular to the plane, which is ( mg \cos(\theta) ), where ( m ) is the mass of the object, ( g ) is the acceleration due to gravity, and ( \theta ) is the angle of the incline.

Given:

  • ( m = 6 ) kg
  • ( g \approx 9.8 ) m/s² (acceleration due to gravity)
  • ( \theta = -\frac{\pi}{6} )

We have: [ N = mg \cos(\theta) = 6 \times 9.8 \times \cos\left(-\frac{\pi}{6}\right) ]

The force of kinetic friction ( F_k ) is given by:

[ F_k = \mu_k \cdot N ]

where ( \mu_k ) is the coefficient of kinetic friction.

Given that it takes 18 N to start pushing the object down the plane and 1 N to keep pushing it, we have:

[ F_s = 18 \text{ N} ] [ F_k = 1 \text{ N} ]

We can now solve for ( \mu_s ) and ( \mu_k ) using the formulas for static and kinetic friction.

[ \mu_s = \frac{F_s}{N} ] [ \mu_k = \frac{F_k}{N} ]

After calculating ( N ), ( \mu_s ), and ( \mu_k ), we'll have the coefficients of static and kinetic friction.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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