Question

A 2.5 kg block is kept on a 3.2 kg block resting on the floor of an elevator. If the elevator is moving up at 1.3 m/s2, calculate the force exerted by the 2.5 kg block on the 3.2 kg block.

Answer 1

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Answer on Question #48629, Physics, Mechanics | Dynamics | Kinematics

A $2.5\,\mathrm{kg}$ block is kept on a $3.2\,\mathrm{kg}$ block resting on the floor of an elevator. If the elevator is moving up at $1.3\,\mathrm{m/s^2}$, calculate the force exerted by the $2.5\,\mathrm{kg}$ block on the $3.2\,\mathrm{kg}$ block.

By the Second Newton’s law (2.5kg):

$$F_{3.2} - mg = ma$$

Where $F_{3.2}$ is a force $3.2\,\mathrm{kg}$ block acts on a $2.5\,\mathrm{kg}$ block

$$F_{3.2} = mg + ma = m(g + a)$$

By the Third Newton’s law $F_{3.2} = F_{2.5}$ (values)

So, the force exerted by the $2.5\,\mathrm{kg}$ block on the $3.2\,\mathrm{kg}$ block:

$$F_{2.5} = m(g + a) = 2.5\,\mathrm{kg} \cdot \left(9.8\,\frac{m}{s^2} + 1.3\,\frac{m}{s^2}\right) \approx 28\,\mathrm{N}$$

Answer: $F_{2.5} \approx 28\,\mathrm{N}$