An object, previously at rest, slides #8 m# down a ramp, with an incline of #pi/4 #, and then slides horizontally on the floor for another #9 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?

Answer 1

The coefficient of kinetic friction is #=0.39#

At the top of the ramp, the object possesses potential energy.

At the bottom of the ramp, part of the potential energy is converted into kinetic energy and the work done by the frictional force.

The ramp is #s=8m#
The angle is #theta=1/4pi#

Therefore,

#PE=mgh=mg*ssintheta#
#mu_k=F_r/N=F_r/(mgcostheta)#
#F_r=mu_k*mgcostheta#
Let the speed at the bottom of the ramp be #ums^-1#

so,

#mg*ssintheta=1/2m u^2+s*mu_k*mgcostheta#
#gssintheta=1/2u^2+smu_kgcostheta#
#u^2=2gs(sintheta-mu_kcostheta)#

On the horizontal part,

The initial velocity is #=u#
The final velocity is #v=0#
The distance is #d=9m#

The deceleration is calculated with Newton's Second Law

#F=F'_r=mu_kmg=ma#
#a=-mu_kg#

We apply the equation of motion

#v^2=u^2+2ad#
#0=2gs(sintheta-mu_kcostheta)-2mu_kgd#
#2mu_kgd=2gssintheta-mu_k2gscostheta#
#2mu_kgd+mu_k2gscostheta=2gssintheta#
#mu_k(2gd+2gscostheta)=2gssintheta#
#mu_k=(gssintheta)/(gd+gscostheta)#
#=(9.8*8*sin(1/4pi))/(9.8*9+9.8*8*cos(1/4pi))#
#=55.44/143.64#
#=0.39#
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Answer 2

To find the coefficient of kinetic friction, you can use the equation:

[ \mu_k = \frac{f_k}{N} ]

Where:

  • ( \mu_k ) is the coefficient of kinetic friction.
  • ( f_k ) is the force of kinetic friction.
  • ( N ) is the normal force.

First, calculate the gravitational force component parallel to the ramp's surface: [ F_{\text{parallel}} = mg \sin(\theta) ] [ F_{\text{parallel}} = (m)(9.8 , \text{m/s}^2) \sin(\pi/4) ] [ F_{\text{parallel}} = (m)(9.8 , \text{m/s}^2) \cdot \frac{\sqrt{2}}{2} ] [ F_{\text{parallel}} = 4.9m ]

Then, find the work done by this force: [ W_{\text{parallel}} = F_{\text{parallel}} \cdot d_{\text{ramp}} ] [ W_{\text{parallel}} = (4.9m) \cdot (8 , \text{m}) ] [ W_{\text{parallel}} = 39.2m ]

Next, find the work done by the kinetic friction on the horizontal surface: [ W_{\text{friction}} = f_k \cdot d_{\text{horizontal}} ] [ W_{\text{friction}} = \mu_k \cdot N \cdot d_{\text{horizontal}} ] [ W_{\text{friction}} = \mu_k \cdot mg \cdot \cos(\theta) \cdot d_{\text{horizontal}} ] [ W_{\text{friction}} = \mu_k \cdot (m)(9.8 , \text{m/s}^2) \cdot \cos(\pi/4) \cdot 9 ] [ W_{\text{friction}} = \mu_k \cdot (m)(9.8 , \text{m/s}^2) \cdot \frac{\sqrt{2}}{2} \cdot 9 ] [ W_{\text{friction}} = 4.9 \mu_k \cdot m \cdot 9 ]

Since energy is conserved, the work done by the parallel force on the ramp should be equal to the work done by kinetic friction on the horizontal surface: [ 39.2m = 4.9 \mu_k \cdot m \cdot 9 ]

Solving for ( \mu_k ): [ \mu_k = \frac{39.2}{4.9 \cdot 9} ] [ \mu_k = \frac{39.2}{44.1} ] [ \mu_k ≈ 0.89 ]

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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.

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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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