Question

A 6.0-kg box moving at 3.0 m/s on a horizontal, frictionless surface runs into a light spring of force constant 75 N/cm. Use the work-energy theorem to find the maximum compression of the spring

Answer 1

Answer:

8.5 cm

Explanation:

According to the work energy theorem, the total work done equals the change in the kinetic energy of the particle.

Therefore,

$W = K_{total}$

where,

$K_{tot} = K_{f} - K_{i}$

Here, W is the total work done by force on the spring.

plug in the following values:

$\\ K_{i} = \frac{mv^{2} }{2} \\ K_{f} = 0\\W = -\frac{kx^{2} }{2}$

We get:

$-\frac{kx^{2} }{2} = 0 -\frac{mv^{2} }{2}$

Solving this

$x = v\sqrt{\frac{m}{k} }$

Plug in the values given in the question:

$x = 3\sqrt{\frac{6}{7500} }$

$x = 8.5 cm$

Answer 2

The maximum compression of the spring will be "0.08 m".

According to the question,

Mass of the box,

• m = 6 kg

Speed of the box,

• V = 3 m/s

Force constant of spring,

• k = 75 N/cm

           = 75 × 100

           = 7500 N/m

By using the conservation of energy, we get

→ $\frac{mV^2}{2} = \frac{kX^2}{2}$

or,

→ $mV^2=kX^2$

By substituting the values, we get

→ $6(3)^2= (7500)X^2$

→      $54= 7500 \ X^2$

      $X^2 = \frac{54}{7500}$

      $X^2= 0.0072$

       $X = 0.08 \ m$ (Max. spring compression)

Thus the above approach is correct.

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