Respuesta :
Answer:
Explanation:
Given
mass of ice cube [tex]m=50 gm[/tex]
inclination [tex]\theta =25^{\circ}[/tex]
Spring constant [tex]k=25\ N/m[/tex]
initial compression [tex]x=0.1 m[/tex]
Suppose Initially the cube is at the datum and it moves h height after release
So Elastic Potential Energy of Spring will convert to Potential Energy of cube
[tex]\frac{1}{2}kx^2=mgh[/tex]
[tex]\frac{1}{2}\times 25\times (0.1)^2=50\times 10^{-3}\times 9.8\times h[/tex]
[tex]h=0.255 m[/tex]
so final Potential Energy will be [tex]=mgh=0.05\times 9.8\times 0.255=0.125 J[/tex]
Initial Potential Energy[tex]=0 J[/tex]
B.Initially spring is at compression so energy associated with it is its Elastic Potential Energy i.e. 0.125 J
Final Energy[tex]=0 J[/tex]
The final potential energy of the object is equal to the initial Elastic Potential Energy of the spring.
When Ice cube released from the spring, Elastic Potential Energy of Spring will convert to Potential Energy of cube,
[tex]\bold {\dfrac 1{2}kx^2 = mgh}[/tex]
Where,
k - Spring constant = 25 N/m
x - initial compression = 0.1 m
m - mass = 50 g
g - gravitational acceleration = 9.8 m/s
h - height
Put the values in the formula and solve it for h
[tex]\bold {\bold {\dfrac 1{2}(25\times 0.1 ^2) = 50\times 9.8 \times h}}\\\\\bold {h = 0.225 m}[/tex]
Thus, final potential energy will be,
[tex]\bold {U_f =0.05 \times 9.5 \times 0.225}\\\\\bold {U_f =0.125 J}[/tex]
Since, initially ice cube is at rest so the initial potential energy will be zero.
B.Initially spring is at compression Elastic Potential Energy will be equal to the final potential energy
Initial Elastic Potential Energy= 0.125 J
Final Elastic Potential Energy = 0 J
Therefore, the final potential energy of the object is equal to the initial Elastic Potential Energy of the spring.
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