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Work Power and Energy question

2007 · Shift 0 · Q83

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Work Power and Energy
2007 · Shift 0 · Q83

Work Power and Energy question

2007 · Shift 0 · Q83

JEE MainPhysicsWork Power and EnergyMCQ+4 / −1

A

2kg

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, 000N/m.

The spring compresses by

A
$8.5cm$
B
$5.5cm$
C
$2.5cm$
D
$11.0cm$

View written solutionFree

Correct answer: B

Given data

Mass of block: $m = 2\,\text{kg}$
Initial speed: $u = 4\,\text{m/s}$
Friction force: $f = 15\,\text{N}$
Spring constant: $k = 10000\,\text{N/m}$
Final speed at maximum compression: $v = 0$

Apply work-energy principle

The initial kinetic energy of the block is spent in:

compressing the spring,
doing work against friction.

So,

\frac{1}{2}mu^2 = \frac{1}{2}kx^2 + fx

Substitute values

Initial kinetic energy:

\frac{1}{2}(2)(4^2) = 16\,\text{J}

Hence,

16 = \frac{1}{2}(10000)x^2 + 15x

16 = 5000x^2 + 15x

Rearranging,

5000x^2 + 15x - 16 = 0

Solve the quadratic equation

Using quadratic formula,

x = \frac{-15 \pm \sqrt{15^2 - 4(5000)(-16)}}{2(5000)}

x = \frac{-15 \pm \sqrt{225 + 320000}}{10000}

x = \frac{-15 \pm \sqrt{320225}}{10000}

Now,

\sqrt{320225} \approx 565.88

Taking the positive root,

x = \frac{-15 + 565.88}{10000} \approx \frac{550.88}{10000}

x \approx 0.0551\,\text{m}

x \approx 5.51\,\text{cm}

Match with options

$5.51\,\text{cm}$ is closest to $5.5\,\text{cm}$ .

Therefore, the correct option is:

\boxed{\text{B}}

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