Frictional oscillations-1

Because the acceleration of the block is in the plane of the slope, we deduce that the normal force has to be constant , and is given by $F_n = mg \cos θ$

The forces that determine the motion are along the slope of the plane:

• the friction force ( $F_f =\pm μ mg \cos θ$ , directed against the motion),
• the spring force ( $F_s=-kx$ ),
• the parallel component of gravitation ( $F_{par}= -mg \sin θ$ )

For a moment, if we ignore friction, we see that gravitation is canceled out by spring force at $x=-14 \text{cm}$ . Setting our origin at -14cm below where the block starts, this compensates for the parallel force, and using the values $k=1, m=2, g=10, \sin θ =\frac{7}{10}, \cos θ = \sqrt{1-\frac{49}{100}} = \frac{\sqrt{51}}{100}$ we now get the differential equations

$2\ddot{x}=1-x\text{ when moving downward}$

$2\ddot{x}=-1-x\text{ when moving upward}$

For the first stroke, using the starting conditions $x(0)=14$ and $\dot{x}(0)=0$ we get $x(t)=1+13 \cos\frac{t}{\sqrt{2}}$

This ends when $\dot{x}(t)=\frac{13}{\sqrt{2}} \sin\frac{t}{\sqrt{2}}=0$ , so at $t_1=\sqrt{2}π$ .

Hence we have $x(t_1)=1-13=-12$

Using the second d.e. and conditions $x(t_1)=-12, \dot{x}(t_1)=0$ we get

$x(t) = 1 + 13 \cos\frac{t}{\sqrt{2}}$ for $t \in [0, \sqrt{2}π]$

$x(t) = -1 + 11 \cos\frac{t}{\sqrt{2}}$ for $t \in [\sqrt{2}π, 2\sqrt{2}π]$

$x(t) = 1 + 9 \cos\frac{t}{\sqrt{2}}$ for $t \in [2\sqrt{2}π, 3\sqrt{2}π]$

$x(t) = -1 + 7 \cos\frac{t}{\sqrt{2}}$ for $t \in [3\sqrt{2}π, 4\sqrt{2}π]$

$x(t) = 1 + 5 \cos\frac{t}{\sqrt{2}}$ for $t \in [4\sqrt{2}π, 5\sqrt{2}π]$

$x(t) = -1 + 3 \cos\frac{t}{\sqrt{2}}$ for $t \in [5\sqrt{2}π, 6\sqrt{2}π]$

$x(t) = 1 +  \cos\frac{t}{\sqrt{2}}$ for $t \in [6\sqrt{2}π, 7\sqrt{2}π]$

Now motion will stop when the block halts at a distance from equilibrium small enough that friction can hold the spring force. Since the friction = $|F_f| \le μ mg \cos θ = 1$ , this maximum distance from equilibrium is 1. So the block stops at -12, +10, -8, +6, -4, +2, and finally at 0.

A. We already saw that the spring was stretched by 14 cm.

B. Total distance traveled=14+12+12+10+10+8+8+6+6+4+4+2 = 98 cm. Energy check using this result: Work done =1×0.98=0.98J; Spring energy =0.5×100×(0.14)^2=0.98 J

C. We count 7 stops.

So the required answer is $14+98+7=119$ .

The increment of length required (A) to satisfy absolute rest when the net force in the motion direction = 0 A - 20 * 7/10 = 0 A=14 cm Acceleration a = net force /m let x be the distance between the object new position and the initial point in (Cms)

a=v.dv/dx= (13 - x)/2 when falling down and = (15 - x)/2 when rising Integrating (a) with respect to x from the current position to the new x = v^2/2

Thus v(down) = sqrt(13x-50x^2-Ci) v(up) = sqrt(15x-50x^2-Cj) Ci & Cj are constants depending on the integration When V = 0 the object will be momentary at rest until it rest at position x=14 the positions that object rest on are : x1= 26 x2= 4 x3= 22 x4= 8 x5= 18 x6= 12 x7= 14 So C = 7 B = (x1-x0) +(x1-x2)+ (x3-x2)+.......+(x7_x6) = 98

So A + B +C = 119