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Question: A block of mass m moving at a speed v compresses a spring through a distance x before its speed is h...

A block of mass m moving at a speed v compresses a spring through a distance x before its speed is halved, the spring constant of the spring is
A. 3vm4x2\dfrac{{3vm}}{{4{x^2}}}
B.3v2m4x2\dfrac{{3{v^2}m}}{{4{x^2}}}
C. 3vm34x3\dfrac{{3v{m^3}}}{{4{x^3}}}
D.4v2m23x2\dfrac{{4{v^2}{m^2}}}{{3{x^2}}}

Explanation

Solution

Use the formula for the kinetic energy of an object. Also use the formula for the spring energy of the spring. Use the law of conservation of energy. According to the law of conservation of the energy, the initial kinetic energy of the block is equal to the final kinetic energy of the block and the spring energy of the spring.

Formulae used:
The kinetic energy KK of an object is
K=12mv2K = \dfrac{1}{2}m{v^2} …… (1)
Here, mm is the mass of the object and vv is the velocity of the object.
The spring energy Es{E_s} is given by
Es=12kx2{E_s} = \dfrac{1}{2}k{x^2} …… (2)
Here, kk is the spring constant and xx is the displacement of the spring.

Complete step by step answer:
We have given that the block of mass mmmoving with speed vvcompresses a spring by xx until its speed is reduced to half of its initial speed.
Hence, the initial speed of the block is vv and the final speed of the block is v2\dfrac{v}{2}.
vi=v{v_i} = v and vf=v2{v_f} = \dfrac{v}{2}
According to the law of conservation of energy, the initial kinetic energy Ki{K_i} of the block is equal to the sum of the final kinetic energy Kf{K_f} of the block and the gain in spring energy Es{E_s}.
Ki=Kf+Es{K_i} = {K_f} + {E_s}
Substitute 12mvi2\dfrac{1}{2}mv_i^2 for Ki{K_i}, 12mvf2\dfrac{1}{2}mv_f^2 for Kf{K_f} and 12kx2\dfrac{1}{2}k{x^2} for Es{E_s} in the above equation.
12mvi2=12mvf2+12kx2\dfrac{1}{2}mv_i^2 = \dfrac{1}{2}mv_f^2 + \dfrac{1}{2}k{x^2}
Substitute vv for vi{v_i} and v2\dfrac{v}{2} for vf{v_f} in the above equation.
12mv2=12m(v2)2+12kx2\dfrac{1}{2}m{v^2} = \dfrac{1}{2}m{\left( {\dfrac{v}{2}} \right)^2} + \dfrac{1}{2}k{x^2}
kx2=mv2mv24\Rightarrow k{x^2} = m{v^2} - m\dfrac{{{v^2}}}{4}
kx2=3mv24\Rightarrow k{x^2} = \dfrac{{3m{v^2}}}{4}
k=3v2m4x2\Rightarrow k = \dfrac{{3{v^2}m}}{{4{x^2}}}
Therefore, the spring constant of the spring is 3v2m4x2\dfrac{{3{v^2}m}}{{4{x^2}}}.

Hence, the correct option is B.

Note:
One can also solve the same question by considering the change in kinetic energy of the block is equal to the change in spring energy of the spring or the change in kinetic energy of the block provides the spring energy to the spring. The change in kinetic energy and the change in spring energy both will be with negative sign and hence the negative sign gets cancelled and the ultimate answer is the same.