Question

1.3 The equation mv gl sin 0 +kl2 describes the energy of an object with mass m amd initial speed vo sliding a distance l up a frictionless ramp at inclination angle θ, all the while compressing a spring with stiffness k as it comes to rest. Solve the equation for I, the distance the object moves up the incline and compresses the spring.
Please be as detailed as possible so I can follow each step to learn this. Thanks!

Answer 1

Solution :

We are given that

$\frac{mv^2_o}{2} = mglsin\theta+\frac{kl^2}{2}$........................(1)

Above equation represents the conservation of mechanical energy and we have to find "l" from this equation.

If we look carefully then we can see that this equation is a quadratic equation in "l".

First, let's rearrange this equation in standard form i.e. in the below form

$ax^2+bx+c=0$

rearranging equation (1),we have

$kl^2+2mglsin\theta-mv^2_o=0$..................(2)

Now solution of a quadratic equation is given by

$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$............(3)

comparing equation (1) and (2)

$x= l, \ a =k ,\ b = 2mgsin\theta,\ c =-mv^2_o$

Now to find "l" put all values in equation (3)

$l= \frac{-2mgsin\theta\pm \sqrt{4m^2g^2sin^2\theta+4kmv^2_o}}{2k}$

So their are two possible values of "l" but one of them is negative so the only possible value of "l" is

$l= \frac{-2mgsin\theta+ \sqrt{4m^2g^2sin^2\theta+4kmv^2_o}}{2k}$......................Answer.