A3 kg block is placed at the top of a track consisting of two frictionless quarter circles of radius r equal to 2 m connected by a 7 m long, straight, horizontal surface. the coefficient of kinetic friction between the block and the horizontal surface is mu k equal to .1. the block is released from rest. from an expression that describes where the block stops, as measured from the left end of the horizontal surface, marked x=0.

The expressions that describes where the block stops from the left end of the horizontal surface, marked x = 0 is;

The reasons why the above values are correct are as follows;

The given parameters are;

Mass of the block, m = 3 kg

Radius of the quarter circular track, R = 2 m

Length of the horizontal track, L = 7 m

Coefficient of friction between the block and the horizontal surface, = 0.1

Required:

To form an expression that describes where the block stops as measured from the left end of the horizontal surface marked x = 0

Solution;

A quarter of a circle has the radii lengths arranged at right angles to each other, both of which are perpendicular to tangents of the circle

A flat surface, such as the horizontal surface in the question is a tangent to the circle

Therefore;

The radius of the circle, perpendicular to the horizontal surface = The height from which the block is released

Potential energy = m·g·h

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Therefore, the potential energy of the block, P.E., is given as follows;

P.E. = m·g·R

P.E. = 3 kg × 2 m × 9.81 m/s² = 58.86 J

The potential energy of the block, P.E. = 58.86 J

The frictional force due to the block, = m·g·

3 kg × 9.81 m/s²× 0.1 = 2.943 N

Work done by the friction as the block moves along the horizontal surface, is given as follows;

Work done = Force × Distance moved in the direction of the force

∴ Work done by friction = × d

Where;

d = The distance from the left end where the block reaches

Work done by friction also = Energy transferred = ΔE

When the block stops, the total energy in the block is transferred

Therefore, when the block stops, according to the law of conservation of energy;

Work done by friction = × d = Energy transferred = ΔE = P.E.

2.943·d = 58.86

Therefore, given that work is done by friction only on the horizontal plane, we have;

The distance travelled on the horizontal surface = 20 m

Therefore, the block traverses the horizontal surface twice, and then for another 6 meters

Let N represent the number of times the block traverses on the horizontal surface, and let L represent the length of the horizontal surface, we have;

d = N·L + x (when N is even)

d = N·L + L - x = L·(N + 1) - x (when N is odd)

Therefore;

The expressions that describe where the block stops as measured from the left end of the horizontal surface marked x = 0 are;

and  

The distance where the block stops from the left end is 6 meters, therefore, we have;

Learn more about the law of conservation of energy here:

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0.93xc=0.93x3x10^8m/s

2.54cm is 1 inch

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