Question

An object of mass m 5 kilograms falls vertically to the ground under the action of the earth gravitational acceleration of magnitude g 10 meters per second squared. Denote by y vertical coordinate, positive upwards, and let y 0 be at the earth surface. Recall that the force on the object in this situation is f--mg, where the negative sign says the force points downwards. (a) Write the differential equation satisfied by this system. y"-10 Note: Write t for t, write y for y(t), and yp for y' (t). (b) Find the mechanical energy E of this system E(t)0y+5/2(p)2 Note: Write t for t, write y for y(t), and yp for y'(t). (c) If the initial position of the object is y(0) -4 meters and its initial velocity is y' (0) 0 meters per second, find the value of the speed of the object l'l when the object reaches the ground. Help Entering Answers

Answer 1

Sol. A :- According to Newton's equation of motion

$S = ut + 1/2at^2$

Here we have S = y and a = g (gravitational pull) = 10m/s^2 and u = 0 ( object falling freely)

So the equation is, $y = 5t^2$

Differentiating y w.r.t. Time y' = 10t

Again differentiating w.r.t. Time we have y" = 10

This is our desired differential equation.

Sol. B :- Mechanical energy of a freely falling object is the sum of potential energy and kinetic energy i.e.

E_M = E_potential + E_kinetic

_{And we know,}

E_potential = mgh and E_kinetic = $1/2mv^2$

As given m = 5kg and g = 10m/s and h = y and v = y' (or yp according to note)

  E_m= 50y + 5/2(y')^2.

This is your required Equation.

Sol. 3 :- Newton's equation of motion for a freely falling object is

$y = ut + 1/2gt^2$ In this case we are given Y= 4m and u = 0

So we have $t = \sqrt{2y/g}$

Also we know that V = u + gt , and where u = 0 , So V = gt

where V is final velocity. Putting value of t from previous equation, we get

V = $\sqrt{2yg}$

so final speed is $\sqrt{80}$ which is equal to 8.9443 m/s .

This is your desired answer.

Hope this helps :-)