Respuesta :
Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=[tex]\frac{1}{2} kx^{2} +PE1[/tex]
Energy at the point where the block will stop consists of only gravitational potential energy=[tex]PE2[/tex]
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒[tex]\frac{1}{2} kx^{2} +PE1=PE2[/tex]
⇒[tex]PE2-PE1=\frac{1}{2} kx^{2}[/tex]
Also [tex]PE2-PE2=mgh[/tex]
where [tex]m[/tex] is the mass of block
[tex]g[/tex] is acceleration due to gravity=9.8 m/s
[tex]h[/tex] is the difference in height between two positions
⇒[tex]mgh=\frac{1}{2} kx^{2}[/tex]
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴[tex]2100*9.8*h=\frac{1}{2}*2200*0.11^{2}[/tex]
⇒[tex]20580*h=13.31[/tex]
⇒[tex]h=\frac{13.31}{20580}[/tex]
⇒h=0.0006467m=[tex]6.5e-4[/tex]
