The movement of the eraser thrown by the student vertically upward is described by the equation

The movement of the eraser thrown by the student vertically upward is described by the equation y = A + Bt + Ct ^ 2, where A = 1.2 m, B = 1 m / s, C = -0.5 m / s2. Determine the mechanical energy of the eraser at the initial moment of time, as well as its kinetic and potential energies after a time interval of t = 0.50 s. Eraser weight m = 8 g.

Since the movement of the eraser thrown by the student vertically upward is described by the equation y = A + B ∙ t + C ∙ t ^ 2, where A = 1.2 m, B = 1 m / s, C = – 0.5 m / s ^ 2, then comparing it with the equation of uniformly accelerated motion of a body, written in general form: y = y₀ + v₀ｙ ∙ t + a ∙ (t ^ 2) / 2, where y₀ is the initial ordinate of the body, v₀ｙ is the projection of the initial speed of movement on the vertical axis, a – acceleration, t – movement time, we get: y₀ = 1.2 m; v₀ｙ = 1 m / s; a / 2 = – 0.5 m / s ^ 2 or a = – 1 m / s ^ 2. Eraser weight m = 8 g = 0.008 kg.

To determine the mechanical energy of the eraser W = Wk + Wp at the initial time t = 0 s, we find its kinetic Wk and potential Wp energy for t = 0 s by the formulas: Wk = m ∙ (v₀ｙ) ^ 2/2 and Wp = m ∙ g ∙ у₀, where the coefficient g = 9.8 N / kg. We get:

W = m ∙ (v₀ｙ) ^ 2/2 + m ∙ g ∙ у₀;

W = 0.008 kg ∙ (1 m / s) ^ 2/2 + 0.008 kg ∙ 9.8 N / kg ∙ 1.2 m;

W ≈ 0.098 J ≈ 98 mJ.

To determine the kinetic energy of the eraser after a time interval t = 0.5 s, we first find the speed vｙ of the eraser by the formula: vｙ = v₀ｙ + a ∙ t or vｙ = 1 m / s – 1 m / s ^ 2 ∙ 0.5 s; vｙ = 0.5 m / s. Then Wk = m ∙ (vｙ) ^ 2/2. We get:

Wk = 0.008 kg ∙ (0.5 m / s) ^ 2/2; Wc = 0.001 J ≈ 1 mJ. The potential energy will be:

Wп = W – Wк; Wp = 98 mJ – 1 mJ; Wp ≈ 97 mJ.

Answer: the mechanical energy of the eraser at the initial moment of time was 98 mJ, the kinetic and potential energies after t = 0.50 s of steel were ≈ 1 mJ and ≈ 97 mJ.