Thank you! A small block of mass 0.0500 kg starts from rest at a height h = 1.00

Question

Thank you!

A small block of mass 0.0500 kg starts from rest at a height h = 1.00 m and slides down a frictionless incline toward a loop-the- loop, as shown in the figure. The loop-the-loop is also frictionless; however, the horizontal track after the loop- the loop is rough, and the coefficient of kinetic friction between the block and rough track is 0.300. The radius of the loop-the-loop is r = 0.400 m. What is the speed of the block just before it reaches the rough track? 1.01 m/s 4.43 m/s 6.75 m/s 12.2 m/s 30.7 m/s What is the magnitude of the normal force on the block when it is at the top of the loop-the-loop? 0N 0.490N 4.90 N 8.85 N 14.4 N How far from the left end of the rough track does the block travel before coming to rest? 0.910m 1.00m 2.25m 3.33m 9.66m

Explanation / Answer

48.

Using conservation of energy

Potential energy at the top = kinetic energy just before the rough track

mgh = (0.5) m v2

v = sqrt (2gh) = sqrt (2 x 9.8 x 1) = 4.43 m/s

49.

Using conservation of energy

kinetic energy at the bottom of loop = potential energy at top of loop + kinetic energy at top

(0.5) m v2 = mg(2r) + (0.5) m v'2

v2 = 2g(2r) + v'2

4.432 = 4 (9.8) (0.4) + v'2

V' = 1.986 m/s

at the top of loop ., force equation is given as

Fn +mg = m v'2 /r

Fn + (0.05 x 9.8) = 0.05 (1.986)2 / 0.4

Fn = 0

50.

a = acceleration = -ug = 0.3 x 9.8 = - 2.94 m/s2

d = stopping distance

V = initial speed = 4.43

stopping distance is given as

d = V2/2a = 4.432 / (2 x 2.94)

d = 3.34 m

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