Question

A 0.5 kg block slides from the point A (see figure) on a horizontal track with an inital speed of 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/m. The part AB of the track is fictionless and the part BC has the coefficients of static and kinetic friction as 0.22 and 0.2 respectively. If the distances AB and BD are 2 m and 2.14 m respectively, then

• A. frictional force on the block when it comes to rest is 0.2 N.
• B. total distance travelled by the block is 4.24 m.
• C. total compression of spring is 2 m.
• D. frictional force of the block is 5.2 N.

Answer 1

Answer: (A), (B)

The correct options are
A frictional force on the block when it comes to rest is 0.2 N.
B total distance travelled by the block is 4.24 m.

The part AB of the track is frictionless. Hence no energy is lost along AB. The 0.5 kg block travels to spring and compresses the spring by an amount x. Energy is stored in the spring.
Initial kinetic energy of block = (Work done against kinetic friction) + (potential energy of spring)



$\frac{1}{2}M{v}^2={\mu }_{k}Mg(2.14+x)+\frac{1}{2}k{x}^2$

or $\frac{1}{2}\times 0.5\times (3{)}^2$

$=0.2\times 0.5\times 10(2.14+x)+\frac{1}{2}\times 2x^2$

$x=0.1m$

The spring gets compressed by 0.1m

$\therefore$ Restoring force exerted by spring on block

$=2\times 0.1=0.2N$

Maximum possible value of static frictional force

$={\mu }_{S}Mg=\left(0.22\right)\left(0.5\right)\left(10\right)=1.1N$

Since this value is greater than restoring force.

$\therefore$ The block stops after compressing the spring by $x$.

$\therefore$ Total distance travelled by block $=AB+BD+x$

$=2+2.14+0.1=4.24m$