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

[jaredogden]

Homework Statement

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°

Homework Equations

PE = mgz
KE = 1/2mv²
F_r = mu_k*N
Energy from F_r = F_rΔx

The Attempt at a Solution

Since the pellet is at rest KE_A = 0J, however PE_A = (.3kg)(9.81m/s²)(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/s²)(0.10718 m)
PE_B = 0.31543 J

KE_B = 1/2(.3 kg)v_B²

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_B²
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..

 

[LawrenceC]

Would you sketch this?

 

[jaredogden]

It is problem 13.70

 

[jhae2.718]

From my reading of the problem, you're supposed to solve for the friction force on the block at the given points.

 

[jaredogden]

Our teacher just posted an announcement that the problem should be a pellet sliding without friction.

The answer is 2.55N, I'm just unsure how to get here..