i really need help on this physic hw...please help me...?

Question

A small block of mass m = 0.50 kg is fired with an initial speed of v0 = 4.5 m/s along a horizontal section of frictionless track, as shown in the top portion of Figure P7.58. The mass then moves along the frictionless semicircular, vertical tracks of radius R = 1.6 m.

(a) Determine the magnitude of the force exerted by the track on the block at points A and B.

track force at A? N

track force at B ? N

(b) The bottom of the track consists of a section (L = 0.40 m) with friction. Determine the coefficient of kinetic friction between the block and that portion of the bottom track if the block just makes it to point C on the first trip. [Hint: If the block just makes it to point C, the force of contact exerted by the track on the block at that point should be zero.]


here the figure...

http://img.photobucket.com/albums/v372/smilytwinkle/p7_58.jpg

i really need help on this...it so complicated...please help..thanks

Edward...i input your answer into the computer and it said that it wrong..both answer are wrong..can you fix it

Answer 1

a)At points A and B the block experiences centripetal acceleration a1 and acceleration due to gravity g.

At point A the only force that is perpendicular to the surface is the centripetal force.

Fc=ma1 where a1=V^2/R so
Fc=mV^2/R=0.50 x (4.5)^2 / 1.6=6.3N track return this favor with -Fc (action=reaction)

At point B the centripetal force is acting downward and so is the weight of the block due to Earth gravity.

Ft=Fc+ W= Fc + mg
Ft= 6.3N + 0.5 x 9.81=11.2 N. again the track responds with - Ft.

b) The block loses part of its kinetic energy to work done against friction and later against gravity and then it stops at point C.

As the block leaves point B it has a certain kinetic energy Ke2 which is a sum of initial kinetic energy and potential energy Pe1

Ke2=Ke1+Pe1 and Pe1=mgh = mg(2R)= 2mgR

The Ke3 is the kinetic energy the block will have when it looses part of Ke2 when it contributes to work done against friction Wf.

Ke3= Ke2- Wf where Wf=uW L

Finally it must have left just enough energy equal to Pe2 to climb the leg to get to point C.
So Pe2= Ke3 and Pe2=mg(2R)= 2mgR

We are ready to solve this problem

Ke2=Ke1+Pe1
Ke3= Ke2- uW L=Pe2
Since
Ke2- uW L=Pe2
Ke1+Pe1 - uW L=Pe2
Ke1-uWL=Pe2-Pe1=0 since Pe1=Pe2

Finally u=Ke1/(W L)
u=(1/2)m V^2/ (mg L)
u= V^2/ (2g L)=(4.5)^2/(2 x 9.81x 0.4)=
u=2.58