Answer:
Explanation:
350 N force stretches the spring by 30 cm
spring constant K = 350 / 0.30 = (350 / 0.3) N / m
To calculate work done by a spring force we proceed as follows
spring force when the spring is stretched by x = Kx
This force is variable so work done by it can be calculated by integration
Work done by it in stretching from x₁ to x₂
W = ∫ F dx
= ∫ Kx dx with limit from x₁ to x ₂
= 1/2 K ( x₂² - x₁² )
Putting the given values of x₁ = 0.50 m , x₂ = 0.8 m
Work done
= 1/2 x (350 / 0.3)x ( 0.80² - 0.50² )
= 227.50 J
Answer:
The answer is B on khan academy
Explanation:
Answer: 0.5 m
Explanation:
Given
Mass of the person is 
Trampoline launches the person into the air up to height of 
Force experience by springs is 
Here, the work done on displacing the springs is equivalent to the Potential energy acquired by the person i.e.
![\Rightarrow F\cdot x=mgh\quad [\text{x=displacement of the trampoline}]\\\\\text{Insert the values}\\\\\Rightarrow x=\dfrac{50\times 9.8\times 2}{1960}\\\\\Rightarrow x=\dfrac{980}{1960}\\\\\Rightarrow x=0.5\ m](https://tex.z-dn.net/?f=%5CRightarrow%20F%5Ccdot%20x%3Dmgh%5Cquad%20%5B%5Ctext%7Bx%3Ddisplacement%20of%20the%20trampoline%7D%5D%5C%5C%5C%5C%5Ctext%7BInsert%20the%20values%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B50%5Ctimes%209.8%5Ctimes%202%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D%5Cdfrac%7B980%7D%7B1960%7D%5C%5C%5C%5C%5CRightarrow%20x%3D0.5%5C%20m)
Answer:
c. V = k Q1 * Q2 / R1 potential energy of Q1 and Q2 separated by R
V2 / V1 = (R1 / R2) = 1/4
V2 = V1 / 4
Answer:
when work is done on the system or heat comes into the system
