Question

An object of mass m is traveling on a horizontal surface. There is a coefficient of kinetic friction µ between the object and the surface. The object has speed v when it reaches x=0 and encounters a spring. The object compresses the spring, stops, and then recoils and travels in the opposite direction. When the object reaches x=0 on its return trip, it stops.
1. Find k, the spring constant.
2. Express k in terms of µ, m, g, and v.

Answer 1

Answer:

$k= \frac{8\mu^2mg^2}{v^2}$

Explanation:

The mass of object is m

The coefficient of friction is μ

the speed of object at x= 0 is v

when the object compress the springs stops so,

The kinetic energy of the spring is equal to and friction work is done

$\frac{1}{2}mv^2 = \frac{1}{2}kx^2+\mu(mg)x$...........

after that it returns from the rest and finally comes to rest after coming to x=0

position.

So, now the potential energy of the spring is equal to the work done by the friction force.

$\frac{1}{2}kx^2= \mu mgx$

$\frac{1}{2}kx= \mu mg$

$x= \frac{2\mu mg}{x}$

substitute the x value in equation 1

mv^2 = kx^2+2\mu (mg)x

$mv^2= k\frac{2\mu mg}{k}+ 2\mu mg \frac{2\mu mg }{k}$

solving this we get

$mv^2= \frac{8m^2\mu^2g^2}{k}$

therefore the force constant

$k= \frac{8\mu^2mg^2}{v^2}$

Answer 2

Spring Constant is the measure of a spring's stiffness. The value of the spring constant k in terms of µ, m, g, and v is $\dfrac{8m \mu ^2 g^2}{v^2 }$.

What is the spring constant?

Spring Constant is the measure of a spring's stiffness.

Given to us

Mass of object = m

The coefficient of friction = μ

The speed of the object at (x= 0) = v

When the object compresses the spring after a distance the spring and the object both stop, therefore, the Kinetic energy of the object transforms to the kinetic energy of the spring during compression and the potential energy when fully compressed, therefore,

The kinetic energy of the object during the compression,

$\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x$

The kinetic energy of the object after complete compression,

$\dfrac{1}{2}kx^2 = \mu m g x\\\\\dfrac{1}{2}kx = \mu m g \\\\x = \dfrac{2 \mu m g}{k}$

Substitute the value of x in the kinetic energy of the object during the compression,

$\dfrac{1}{2} mv^2 = \dfrac{1}{2}kx^2 + \mu (mg)x\\\\\dfrac{1}{2} mv^2 = \dfrac{1}{2}k(\dfrac{2 \mu m g}{k})^2 + \mu (mg)(\dfrac{2 \mu m g}{k})\\\\\\mv^2 = \dfrac{8m^2 \mu ^2 g^2}{k}\\\\k= \dfrac{8m^2 \mu ^2 g^2}{mv^2 }\\\\k= \dfrac{8m \mu ^2 g^2}{v^2 }$

Hence, the value of the spring constant k in terms of µ, m, g, and v is $\dfrac{8m \mu ^2 g^2}{v^2 }$.

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