Frictional Track
0.5 kg block slides from the points A on a horizontal track with an initial speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant of 2 N/m. The part AB of the track is frictionless and the part BC has coefficient of static and kinetic friction as 0.22 and 0. 2 respectively. Find the total distance through which the block movez before it comes to rest. Length of AB is 2 m and that of BD is 2.14 m.
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1 solution
The block moves without losing any speed on AB. On BD the block faces kinetic friction of magnitude Fk=(0.2)(0.5)(9.8)N=0.98N So,it delelerates at =(0.98÷0.5)m/s^2 = 1.96 m/s^2 At point d, the speed of the block is v=sqrt(3^2 - 2(1.96)(2.14))m/s=0.782 m/s
At that point, the kinetic energy is, Ek=(0.5)(0.5)(0.782)^2 J = 0.1528 J
If the block momenterily stops at a distance 'x' from point D, then- 0.5kx^2 + 0.2*mgx= 0.1528 [conservation of energy] solving the equation gives x=0.14 m
At that time spring force Fs=0.14k=0.28 N But static frictional force Ff=(0.22mg)= 1.078N Ff>Fs ; So the block totally stops there...
The total distance is = (2+2.14+0.14)m =4.28 m = 4.2 m (almost)