Solution to problem 15.5.8 from the collection of Kepe O.E. 1989

15.5.8. Plate 1 with a mass of 40 kg moves translationally and rectilinearly with a speed v1 = 1 m/s. Body 2 with a mass of 10 kg moves translationally with respect to the plate at a speed vr = 0.4 m/s. Determine the kinetic energy of the system of bodies if the vectors v1 and vr are parallel.

To solve the problem, we use the formula for kinetic energy:

$$E_{\text{к}} = \frac{m_1 v_1^2}{2} + \frac{m_2 v_2^2}{2},$$

where $m_1$ and $m_2$ are the masses of bodies, $v_1$ and $v_2$ are the velocities of bodies.

Considering that the vectors $v_1$ and $v_r$ are parallel, the speed of body 2 relative to the plate is equal to:

$$v_2 = v_1 + v_r.$$

Substituting the expression for $v_2$ into the formula for kinetic energy, we obtain:

$$E_{\text{к}} = \frac{m_1 v_1^2}{2} + \frac{m_2 (v_1 + v_r)^2}{2}.$$

Substituting numerical values, we get:

$$E_{\text{к}} = \frac{40 \cdot 1^2}{2} + \frac{10 \cdot (1 + 0,4)^2}{2} \approx 24,8 \,\text{Дж}.$$

Thus, the kinetic energy of the system of bodies is approximately 24.8 J.

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solution to problem 15.5.8 from the collection of Kepe O.E. 1989.

In this problem, we have a plate with a mass of 40 kg, moving translationally and rectilinearly with a speed v1 = 1 m/s. There is also a body with a mass of 10 kg, moving translationally relative to the plate with a speed vr = 0.4 m/s. Vectors v1 and vr are parallel. It is necessary to determine the kinetic energy of the system of bodies.

After payment, you will receive a solution to the Kepe problem No. 15.5.8 from the solution book for the collection of short problems on theoretical mechanics by O.E. Kepe. The task is completed in Word format (saved as an image in PNG format) and will open on any PC, smartphone or tablet. After checking the solution, we will be very grateful if you leave positive feedback.

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