# A box with an initial speed of #5 m/s# is moving up a ramp. The ramp has a kinetic friction coefficient of #5/4 # and an incline of #(3 pi )/4 #. How far along the ramp will the box go?

The box will go

I used a combination of kinematics and Newton's second law.

#|->v_i=5"m"//"s"# #|->mu_k=5/4# #|->theta=(3pi)/4->45^o# #|->g=9.81"m"//"s"^2#

Diagram:

Note no acceleration in the vertical direction (dynamic equilibrium).

Therefore, we have

The force of kinetic friction is given by

#color(darkblue)(f_k=mu_kn)#

We can find the perpendicular component of gravity and consequently the magnitude of the normal force using trigonometry. We can see from the above diagram that the angle between the force of gravity and the vertical is the same as the angle of incline of the ramp.

#cos(theta)="adjacent"/"hypotenuse"#

#=>cos(theta)=(F_G)_y/(F_G)#

#=>color(darkblue)((F_G)_y=F_Gcos(theta))#

Similarly, we find that

As

#color(darkblue)(n=mgcos(theta))#

Therefore:

#color(darkblue)(f_k=mu_k*mgcos(theta))#

Substituting this and the equation we derived for the force of gravity parallel above into our statement for the net force parallel:

#mu_kcancelcolor(skyblue)(m)gcos(theta)+cancelcolor(skyblue)(m)gsin(theta)=cancelcolor(skyblue)(m)a_x#

#=>color(blue)(a_x=g(mu_kcos(theta)+sin(theta)))#

We can now use this kinematic equation to find the distance traveled up the ramp:

#v_f^2=v_i^2+2aDeltas#

Solving for

#Deltas=(cancel(v_f^2)-v_i^2)/(2a)#

Substituting in the equation we derived above for the acceleration:

#color(blue)(Deltas=-v_i^2/(2g(mu_kcos(theta)+sin(theta)))#

Now we can substitute in our known values:

#Deltas=-(5"m"//"s")^2/(2(9.81"m"//"s"^2)(5/4cos(45^o)+sin(45^o))#

#=>color(blue)(~~-0.80"m")#

Because we defined *up the ramp* as negative, getting a negative value for displacement tells us that our final position was *less* than our initial position, which makes sense as we set at

- Note: this question can also be solved using the work-energy theorem, but it can be very easy to misplace negative signs and alter your answer since addition is involved in the denominator. Here is a link to an example of a very similar problem solved using the W-E theorem.

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To determine how far along the ramp the box will go, we need to calculate the distance traveled using the equations of motion and the forces acting on the box. The kinetic friction force ( f_k ) can be calculated using the formula ( f_k = \mu_k \cdot N ), where ( \mu_k ) is the coefficient of kinetic friction and ( N ) is the normal force. The normal force can be determined by resolving the weight of the box into components perpendicular and parallel to the incline. Once we have the friction force, we can use Newton's second law to calculate the acceleration of the box along the ramp. With the initial speed and acceleration, we can then use the equations of motion to find the distance traveled.

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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.

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.

- An object with a mass of #10 kg# is on a plane with an incline of # - pi/4 #. If it takes #12 N# to start pushing the object down the plane and #4 N# to keep pushing it, what are the coefficients of static and kinetic friction?
- If an object with a mass of #10 kg # is moving on a surface at #15 ms^-1# and slows to a halt after # 4 s#, what is the friction coefficient of the surface?
- An object, previously at rest, slides #5 m# down a ramp, with an incline of #(3pi)/8 #, and then slides horizontally on the floor for another #15 m#. If the ramp and floor are made of the same material, what is the material's kinetic friction coefficient?
- An object with a mass of #5 # # kg# is on a surface with a kinetic friction coefficient of # 4 #. How much force is necessary to accelerate the object horizontally at # 6# #ms^-2#?
- If an object with a mass of #10 kg # is moving on a surface at #45 m/s# and slows to a halt after # 6 s#, what is the friction coefficient of the surface?

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