(a) At the point D
ma=mg
mv²/r=mg
v=sqrt(g•r)=sqrt(9.8•2.87)=5.3 m/s
(b)
KE(C)=PE+KE=
=mg•2r+mv²/2=
=m(2gr+5.3²/2)=54.8 J.
(c)
KE(B)=KE(C) + W(fr)=
= KE(C) + μ•mg•s=
=54.8+0.29•0.78•9.8•4.08=
=63.8 J
(d)
PE1=KE(B)
mgh= KE(B)
h=KE(B)/mg=63.8/0.78•9.8 =8.35 m
A block of mass 0.78kg starts from rest at point A and slides down a frictionless hill of height h. At the bottom of the hill it slides across a horizontal piece of track where the coefficient of kinetic friction is 0.29. This section (from points B to C) is 4.08m in length. The block then enters a frictionless loop of radius r= 2.87m. Point D is the highest point in the loop. The loop has a total height of 2r. Note that the drawing below is not to scale.
a)What is the minimum speed of the block at point D that still allows the block to complete the loop without leaving the track?
b)What is the minimum kinetic energy for the block at point C in order to have enough speed at point D that the block will not leave the track?
c)What is the minimum kinetic energy for the block at point B in order to have enough speed at point D that the block will not leave the track?
d)What is the minimum height from which the block should start in order to have enough speed at point D that the block will not leave the track?
1 answer
1 answer