A $$2\,kg$$ block slides on a horizontal floor with a speed of $$4\,m/s.$$ It strikes a uncompressed spring, and compresses it till the block is motionless. The kinetic friction force is $$15\,N$$ and spring constant is $$10,000 \,N/m.$$ The spring compresses by-
A.
$$8.5 \,cm$$
B.
$$5.5 \,cm$$
C.
$$2.5 \,cm$$
D.
$$11.0 \,cm$$
Answer :
$$5.5 \,cm$$
Solution :
Let the block compress the spring by $$x$$ before stopping.
kinetic energy of the block $$=$$ (P.E. of compressed spring) $$+$$ work done against friction.
$$\eqalign{
& \frac{1}{2} \times 2 \times {\left( 4 \right)^2} = \frac{1}{2} \times 10,000 \times {x^2} + 15 \times x \cr
& 10,000{x^2} + 30x - 32 = 0 \cr
& \Rightarrow 5000{x^2} + 15x - 16 = 0 \cr
& \therefore x = \frac{{ - 15 \pm \sqrt {{{\left( {15} \right)}^2} - 4 \times \left( {5000} \right)\left( { - 16} \right)} }}{{2 \times 5000}} \cr
& \Rightarrow x = 0.055\,m \cr
& \Rightarrow x = \,5.5\,cm \cr} $$
Releted MCQ Question on Basic Physics >> Work Energy and Power
Releted Question 1
If a machine is lubricated with oil-
A.
the mechanical advantage of the machine increases.
B.
the mechanical efficiency of the machine increases.
C.
both its mechanical advantage and efficiency increase.
D.
its efficiency increases, but its mechanical advantage decreases.
A particle of mass $$m$$ is moving in a circular path of constant radius $$r$$ such that its centripetal acceleration $${a_c}$$ is varying with time $$t$$ as $${a_c} = {k^2}r{t^2}$$ where $$k$$ is a constant. The power delivered to the particles by the force acting on it is:
A.
$$2\pi m{k^2}{r^2}t$$
B.
$$m{k^2}{r^2}t$$
C.
$$\frac{{\left( {m{k^4}{r^2}{t^5}} \right)}}{3}$$
A spring of force-constant $$k$$ is cut into two pieces such that one piece is double the length of the other. Then the long piece will have a force-constant of-