Question

The system is initially moving with the cable taut, the 15-kg block moving down the rough incline with a speed of 0.080 m/s, and the spring stretched 39 mm. By the method of this article, (a) determine the velocity v of the block after it has traveled 99 mm, and (b) calculate the distance d traveled by the block before it comes to rest.

Answer 1

Solution :

The spring is expanded by 2 times of the block when it moves down an inclined by x times.

Here, $x_1$ = 39 mm

        $x_2$ = 225 mm

a). From the work energy principal,

   Work forces = kinetic energy

$(mg \sin 50^\circ)\times \frac{99}{1000}-(\mu_k mg \cos 50^\circ) \times \frac{99}{1000} -\frac{1}{2}k(0.225^2 - 0.039^2)=\frac{1}{2}m(V^2_2-0.08^2)$

$(112.6 \times 0.099)-(14.17 \times 0.099)-4.91= 7.5(V^2_2-0.08^2)$

$9.75= 7.5(V^2_2-0.08^2)$

$1.3= V^2_2-0.08^2$

$V_2=1.14\ m/s$

b). calculating the distance travelled by the block before it comes to rest.

Substitute the value of $V_2$ in (1),

$-(\mu_kmg \cos 50^\circ)x + (mg \sin 50^\circ)x-\frac{1}{2}k\left( ( 2x+0.039)^2 - 0.039^2\right)= -\frac{1}{2}m(0.08)^2$

$-14.17x+112.6x - 100(4x^2+0.156x)=-0.048$

$98.43x - 100(4x^2+0.156x)+0.048=0$

$98.43x - 400x^2-15.6x+0.048=0$

$82.83x - 400x^2+0.048=0$

$400x^2- 82.83x-0.048=0$

x = 0.20 m