Answer :
Answer:
Part a)
[tex]L = 0.68 m[/tex]
Part b)
[tex]L = 0.63 m[/tex]
Explanation:
Part a)
As we know that there is no friction in the path
So here we can use energy conservation to find the distance moved by the mass
Initial spring energy = final gravitational potential energy
so we will have
[tex]\frac{1}{2}kx^2 = mgL sin\theta[/tex]
[tex]\frac{1}{2}70(0.5)^2 = 2(9.81)(L) sin41[/tex]
[tex]8.75 = 12.87 L[/tex]
[tex]L = 0.68 m[/tex]
Part b)
Now if spring is connected to the block then again we can use energy conservation
so we will have
[tex]\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2[/tex]
so we will have
[tex]\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2[/tex]
[tex]8.75 = 6.43 + 12.87 x' + 35 x'^2[/tex]
[tex]x' = 0.13 m[/tex]
so total distance moved upwards is
[tex]L = 0.5 + 0.13 = 0.63 m[/tex]
If the spring is not attached, the mass will move 0.68 meters. however, if the spring is attached, the mass will move by 0.63 meters.
How can we arrive at these results?
- We will first calculate the motion of the mass if it is not attached to the spring. This will be done with the following formula:
[tex]\frac{1}{2} kx^2=m*g*L*sin41[/tex]
Within this formula, the symbol "k" represents 70 N/m. The "x" represents 0.50m. The "m" is 2.0 kg, the "g" represents gravity and the L is the value we need to find.
Therefore, we can substitute the values of the formula as follows:
[tex]\frac{1}{2}70(0.50)^2= 2*9.81 *L*sin41\\8.75=12.87L\\L= \frac{8.75}{12.87} = 0.67 ------- 0.68m[/tex]
- Then we can calculate the distance traveled by the mass attached to the spring. For this, we will use the formula:
[tex]\frac{1}{2} kx^2= m*g*(x+x')*sin41+\frac{1}{2} 70x'^2[/tex]
By substituting the values, we can answer the equation as follows:
[tex]\frac{1}{2} 70*0.50^2=2*9.81*(0.50+x')*sin41+\frac{1}{2} 70x'^2\\8.75=6.43+12.87x'+35x'\\X'=0.13m \\\\L=0.5+0.13=0.63 m[/tex]
More information about gravity in the link:
https://brainly.com/question/1479537