(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

This is problem 13.70 from Beer and Johnston Vector Mechanics 9th ed.

A 300-g pellet is released from rest at A and slides with friction

along the surface shown. Determine the force exerted on the pellet

by the surface (a) just before the pellet reaches B, (b) immediately

after it has passed through B.

The drawing shows a pellet A on a declining slope, distance 1.2 m to B which is almost at the bottom of the slope. C is a flat surface. The radius of curvature between B and C is 0.8 m and the angle between them is 30°

2. Relevant equations

PE = mgz

KE = 1/2mv^{2}

F_{r}= mu_{k}*N

Energy from F_{r}= F_{r}Δx

3. The attempt at a solution

Since the pellet is at rest KE_{A}= 0J, however PE_{A}= (.3kg)(9.81m/s^{2})(1.2sin(30°))

(The angle between the horizontal and the 1.2m hypotenuse isn't given, but I just found it using opposite interior angles.)

so PE_{A}= 1.7658 J

At B we will have KE and PE. Assuming C is at z = 0m we can use the radius of curvature and angle given to find z_{b}

z_{B}= 0.8 m - 0.8 m * cos(30°)

z_{B}= 0.10718 m

so PE_{B}= (.3kg)(9.81m/s^{2})(0.10718 m)

PE_{B}= 0.31543 J

KE_{B}= 1/2(.3 kg)v_{B}^{2}

Plugging all of this into the law of conservation of energy where E_{A}= E_{B}

1.7658 J = 0.31543 J + (1/2)(.3 kg)v_{B}^{2}

v_{B}= 3.109 m/s

I got hung up here because I think that the work due to the frictional force should be subtracted from KE_{B}+ PE_{B}in order to get the correct velocity the pellet would be traveling at. However, there is no coefficient of friction given in the problem.

I feel like I'm just missing something small here and my brain is just dead..

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# Dynamics: Finding force exerted on a box traveling down a slope

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