The body, thrown vertically upward, fell 4 seconds after the start of the movement. body weight 200 g.

| February 19, 2021 | education

The body, thrown vertically upward, fell 4 seconds after the start of the movement. body weight 200 g. What is the kinetic energy at the moment of falling? what is the potential energy at the top of the trajectory?

Task data: t (total travel time) = 4 s; m (body weight) = 200 g = 0.2 kg.

Constants: g (acceleration due to gravity) ≈ 10 m / s2.

1) The duration of the fall (rise): t2 = t1 = t / 2 = 4/2 s.

2) Speed ​​at the moment of falling (initial throw speed): Vк = g * t1 = 10 * 2 = 20 m / s.

3) Kinetic energy (we consider the moment of fall): Ek = m * Vk ^ 2/2 = 0.2 * 20 ^ 2/2 = 40 J.

4) Potential energy (consider the upper point of the trajectory): Ep = Ek (energy conservation law) = 40 J.

Answer: The value of kinetic energy at the moment of falling corresponds to the value of potential energy at the highest point of the trajectory and is 40 J.



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