Question

A $2kg$ block slides on a horizontal floor with a speed of $4m/s$. It strikes an uncompressed spring and compresses it until the block becomes motionless. The kinetic friction force is $15N$ and spring constant is $10000N/m$. The spring compresses by:
A. $5.5cm$
B. $2.5cm$
C. $11.0cm$
D. $8.5cm$

Answer 1

Hint: Here, take that the block is already attached to the string, and the frictional force is intact as long as the string compresses, and then apply the law of conservation energy. Hence the distance travelled by the block during the frictional force is the same as the distance of compression of the spring.

Complete step by step answer:
In this case, let the mass of the block be $m=2kg$. The velocity of the block is $4m/s$ which we will denote as v. The spring constant is written as k and its value is $k=10000N/m$. Let the applied frictional force be f and its value is 15N. Let us take the distance that the spring compresses as x meters. As it is not mentioned in the question, we will assume that the frictional force is applied only till the time until the spring compresses, that is it will travel the distance which is the same as that to the compression of the string.

Now, to find the displacement of the string, we will use the law of conservation of energy. The law of conservation of energy states that energy cannot be created nor can it be destroyed. It can only be transferred from one form into another. In our case, the kinetic energy of the block is converted into the potential energy of the spring. The potential energy U of the spring is denoted by U and its value is
\[U = \dfrac{1}{2}k{x^2}\]
Now, applying the law of conservation of energy, we get
\[
\dfrac{1}{2}m{v^2} - fx = \dfrac{1}{2}k{x^2} \\
\Rightarrow \dfrac{1}{2}k{x^2} = 16 - 15x \\
\Rightarrow 5000{x^2} + 15x - 16 = 0 \\
\Rightarrow x = 0.055m \\
\therefore x = 5.5cm \\
\]
Hence option A is the correct option.

Note:To solve the above quadratic equation, one must first find the discriminant and then use the formula to find the roots, of which you will get 0.055 as one of the roots; any other way to solve it will be more time consuming. Here, we only need to balance the energy of the block and the spring in the initial and final position, and not at every position of the trajectory.