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

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

`$(mg \sin 50^\circ)* (99)/(1000)-(\mu_k mg \cos 50^\circ) * (99)/(1000) -(1)/(2)k(0.225^2 - 0.039^2)=(1)/(2)m(V^2_2-0.08^2)$`

`$(112.6 * 0.099)-(14.17 * 0.099)-4.91= 7.5(V^2_2-0.08^2)$`

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

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

`$V_2=1.14\ m/s$`

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

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

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

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

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

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

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

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

x = 0.20 m