Solve Distance for Box A Sliding Down Ramp

In summary, a 30 lb box released from rest slides down a smooth ramp with a coefficient of kinetic friction of 0.6. Using the work-energy equation and the fact that the velocity of the object is independent of its mass, we can calculate that the box will reach a velocity of 16.05 ft/s and travel a distance of 6.67 ft before stopping. However, the actual distance from the end of the platform to where the box stops is 3.33 ft.
  • #1
VinnyCee
489
0
The 30 lb box A is released from rest and slides down along the smooth ramp and onto the surface. Determine the distance s from the end of the surface to where the box stops. The coefficient of kinetic friction between the cart and the box is [itex]\mu_k\,=\,0.6[/itex].

http://img224.imageshack.us/img224/8509/problem14343jq.jpg [Broken]


Here is what I have so far:


[tex]-W\,\Delta\,y\,=\,(-30\,lb)\,(-4\,ft)\,=\,120\,ft\,lb[/tex]

[tex]\sum\,F_y\,=\,N\,-\,W\,=\,0\,\Rightarrow\,N\,=\,W\,=\,30\,lb[/tex]

[tex]\sum\,F_x\,=\,-f_k\,=\,m\,a_x\,\Rightarrow\,-\mu_k\,N\,=\,m\,a_x[/tex]

[tex](-0.6)\,(30\,lb)\,=\,(0.932)\,a_x[/tex]

[tex]a_x\,=\,\frac{-18.6}{0.932}\,=\,-19.3\,\frac{ft}{s^2}[/tex]

Now what?

I know I need to find [itex]v_f[/itex] and the bottom of the hill and I am probably supposed to use a work-energy equation?

[tex]\sum\,T_1\,+\,\sum\,U_{1\,-\,2}\,=\,\sum\,T_2[/tex]

Please help, thanks.
 
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  • #2
VinnyCee said:
know I need to find vf and the bottom of the hill and I am probably supposed to use a work-energy equation?
Yes, as the ramp is frictionless the kinetic energy gained by the block will equal the work done by gravity; 1/2mv2 = mgh. A good point to note for future reference is that the velocity of the object is independent of the mass.
 
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  • #3
Using that, I get this:

[tex]V_f^2\,=\,2\,g\,h\,=\,2\,(32.2)\,(4)\,=\,257.6[/tex]

[tex]V_f\,=\,\sqrt{257.6}\,=\,16.05\,\frac{ft}{s}[/tex]

[tex]v\,=\,v_0\,+\,a\,t[/tex]

[tex]0\,=\,16.05\,+(-19.3)\,t[/tex]

[tex]t\,=\,0.832\,s[/tex]

[tex]s\,=\,s_0\,+\,v_0\,t\,+\,\frac{1}{2}\,a\,t^2[/tex]

[tex]s\,=\,0\,+\,(16.05)\,(0.832)\,+\,\frac{1}{2}\,a\,t^2[/tex]

[tex]s\,=\,6.67\,ft[/tex]

The real answer is 3.33 ft though! What did I do wrong?
 
  • #4
Your going to kick yourself for this one. You have calculated the distance travelled, not the distance from the end of the platform. HINT: What does 10 - 6.67 equal? :wink:
 
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What is the formula for solving distance for box A sliding down a ramp?

The formula for solving distance for box A sliding down a ramp is:
Distance = Height / Sin(Angle), where height is the height of the ramp and angle is the angle of the ramp in degrees.

What are the key factors that affect the distance of box A sliding down a ramp?

The key factors that affect the distance of box A sliding down a ramp are the height of the ramp, the angle of the ramp, and the initial velocity of the box.

How does the mass of box A affect its distance when sliding down a ramp?

The mass of box A does not affect its distance when sliding down a ramp. The distance is determined by the height and angle of the ramp, and the initial velocity of the box. The mass only affects the speed at which the box slides down the ramp.

Can the distance of box A sliding down a ramp be greater than the height of the ramp?

Yes, the distance of box A sliding down a ramp can be greater than the height of the ramp. This is possible when the angle of the ramp is greater than 90 degrees, also known as an overhang ramp. In this case, the distance would be calculated using the formula: Distance = Height x Cos(Angle - 90).

How can the distance of box A sliding down a ramp be increased?

The distance of box A sliding down a ramp can be increased by increasing the height of the ramp or by increasing the angle of the ramp. Another way to increase the distance is by giving the box a higher initial velocity. However, this could also increase the speed at which the box slides down the ramp, so caution should be taken when adjusting the initial velocity.

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