JEE MAIN - Physics (2007 - No. 21)
A $$2$$ $$kg$$ block slides on a horizontal floor with a speed of $$4m/s.$$ It strikes a uncompressed spring, and compress it till the block is motionless. The kinetic friction force is $$15N$$ and spring constant is $$10, 000$$ $$N/m.$$ The spring compresses by
$$8.5cm$$
$$5.5cm$$
$$2.5cm$$
$$11.0cm$$
Explanation
Let the block compress the spring by $$x$$ before coming to rest.
Initial kinetic energy of the block $$=$$ (potential energy of compressed spring) $$+$$ work done due to friction.
$${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$$
$$10,000{x^2} + 30x - 32 = 0$$
$$ \Rightarrow 5000{x^2} + 15x - 16 = 0$$
$$\therefore$$ $$x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$$
$$\,\,\,\,\, = 0.055m = 5.5cm.$$
Initial kinetic energy of the block $$=$$ (potential energy of compressed spring) $$+$$ work done due to friction.
$${1 \over 2} \times 2 \times {\left( 4 \right)^2} = {1 \over 2} \times 10000 \times {x^2} + 15 \times x$$
$$10,000{x^2} + 30x - 32 = 0$$
$$ \Rightarrow 5000{x^2} + 15x - 16 = 0$$
$$\therefore$$ $$x = {{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} } \over {2 \times 5000}}$$
$$\,\,\,\,\, = 0.055m = 5.5cm.$$
Comments (0)
