The stretched spring, contracting, carries with it a body weighing 800 g along the horizontal plane without

The stretched spring, contracting, carries with it a body weighing 800 g along the horizontal plane without friction. At the moment when the deformation of the spring is equal to zero, the velocity of the body is equal to 2 m / s. Determine the amount of spring deformation if its stiffness is 8 kN / m.

To determine the initial value of the spring deformation, we apply the law of conservation of energy: Ek = Ep and m * V ^ 2/2 = k * Δx ^ 2/2, whence Δx = √ (m * V ^ 2 / k).

The values of the variables: m – body weight (m = 800 g; in the SI system m = 0.8 kg); V is the maximum speed of the body (V = 2 m / s); k is the stiffness of the spring (k = 8 kN / m = 8000 N / m).

Let’s make a calculation: Δx = √ (m * V ^ 2 / k) = √ (0.8 * 2 ^ 2/8000) = 0.02 m.

Answer: The spring was stretched 0.02 m.