Question

A $0.5kg$ block slides from a point $A$ (see Fig) on a horizontal track with an initial speed of $3m{s}^{-1}$ towards a weightless horizontal spring of length $1m$ and force constant $2N{m}^{-1}$ . The part $AB$ of the track is frictionless and the part $BC$ has the coefficient of static and kinetic friction as $0.22$ and $0.2$ respectively. If the distance $AB$ and $BD$ are $2m$ and $2.14m$ respectively, find the total distance through which the block moves before it completely stops. (Take $g=10m{s}^{-2}$ ).

• A. 3.23 m
• B. 4.14 m
• C. 4.24 m
• D. 4.67 m

Answer 1

Answer: (C)

As the track AB is frictionless, the block moves this distance = 2m without loss in its initial KE
$=\frac{1}{2}m{v}^2=\frac{1}{2}\times 0.5\times 3^2=2.25J.$
In the path BD (= 2.14 m) as friction is present, so work done against friction = μ_K mgs
= 0.2 × 0.5 × 10 × 2.14 = 2.14 J.
So at D the KE of the block is = 2.25 - 2.14 = 0.11 J.
Now if the spring is compressed by x,
$0.11=\frac{1}{2}\times k\times x^2+{\mu }_{K}mgx$
$i.e.,0.11=\frac{1}{2}\times 2\times x^2+0.2\times 0.5\times 10x$
or x² + x - 0.11 = 0
which on solving gives x = 0.1 m (as x = - 1.1 is inadmissible). After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force
F = kx = 2 × 0.1 = 0.2 N
on the block while limiting frictional force between block and track f_L = μ_s mg = 0.22 × 0.5 × 10 = 1.1 N. Since, F < f_L, the block will not move back. So, the total distance moved by the block
= 2 + 2.14 + 0.1 = 4.24 m