Question

A 3 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.

Answer 1

Answer:

6m

Explanation:

The initial potential energy of the block P.E =mgh

where mass m = 3kg

g= 9.81 m/s2

h = 2m

Hence P.E = 3*9.81*2 = 58.8 Joules

This energy converts to kinetic energy when the block is released from the top

K.E = $mv^2/2 = 58.8$

Solving for v, we obtain 6.26 m/s

The acceleration of the block by Newton's second law of motion,

ma = (μk)*mg

where μk = 0.1

Hence, a = 0.1*9.81 = 0.98m/s/s

Applying Newton's second equation of motion to solve,

$v^2 - u^2 =2as$

where u = 0 m/s (released from rest0

Hence, $6.26^2 = 2*0.98*s$

Solving for s,

s= 20m

Therefore, the distance after two complete cycles = 20 -2(7) = 6m