Respuesta :
Answer:
a) The block moves down a distance of 35 cm down the incline from its rest position to this stopping point
b) The speed of the block just as it touches the spring is 1.7 m/s
Explanation:
First let us determine the spring constant
[tex]k\Delta x=F[/tex]
Given:[tex]F=270N[/tex] and [tex]\Delta x=2cm=0.02m[/tex]
⇒[tex]k=\frac{F}{\Delta x}=\frac{270}{0.02}=13500N/m[/tex]
Let the distance the block moves down the incline from its rest position to this stopping point be L measured along the incline
and height of the block be H from the stopping point
a)Given that [tex]\theta=30[/tex]° and [tex]x=5.5cm=0.055m[/tex] where x is the distance by which the spring compresses
⇒[tex]sin\theta=\frac{H}{L}[/tex]
and using law of conservation of energy:
Energy at rest point=Energy at stopping point
⇒[tex]mgH=\frac{1}{2}kx^2[/tex]
⇒[tex]H=\frac{kx^{2}}{2mg}=\frac{13500*0.055^{2}}{2*12*4.8}=0.174m[/tex]
Therefore, [tex]L=\frac{H}{sin\theta}[/tex]
⇒[tex]L=\frac{0.174}{sin30}=\frac{0.174}{\frac{1}{2}}=0.348m[/tex]
Therefore the block moves down a distance of 35 cm down the incline from its rest position to this stopping point
b) [tex]L=x+l[/tex] where [tex]x[/tex] is the compression in spring and [tex]l[/tex] is the distance between starting of spring to the block
∴[tex]l=L-x=0.348-0.055=0.293m[/tex]
Let h be the height of the block from the starting of spring
∴[tex]h=lsin30=0.293*\frac{1}{2}=0.1465m[/tex]
using law of conservation of energy:
Energy at rest point=Energy at starting point of the spring
⇒[tex]mgh=\frac{1}{2}mv^{2}[/tex] where [tex]v[/tex] is the speed of the block
⇒[tex]v=\sqrt{2gh}=\sqrt{2*9.8*0.1465}=1.7ms^{-1}[/tex]
