Question

A 20 kg block on an inclined plane is pulled up the plane with a rope tied to the block. The rope is at angle of 37 degrees above the surface of the plane. The tension in the rope is 250 N and the frictional force on the block is 8.0 N.What is the acceleration of the block?

Answer 1

Answer on Question #40806 - Physics - Mechanics

A $20\mathrm{kg}$ block on an inclined plane is pulled up the plane with a rope tied to the block. The rope is at angle of 37 degrees above the surface of the plane. The tension in the rope is $250\mathrm{N}$ and the frictional force on the block is $8.0\mathrm{N}$ . What is the acceleration of the block?

Solution

We have, according the second Newton's law that the

$$m = 2 0 k g$$ $$\alpha = 3 7 {}^ {\circ}$$ $$F = 2 5 0 N$$ $$F _ {r} = 8. 0 N$$ $$m \ddot {a} = \vec {F} _ {r} + \vec {P} + m \ddot {g} + \vec {F}$$

Where $\vec{F}_r$ is frictional force, $\vec{P}$ is reaction force of floor, $m\ddot{g}$ is gravitational force, $\vec{F}$ is force applied at an angle of 37 degrees above the horizontal pulls

In terms of projections on directions, which are parallel and perpendicular to plane, have

$$\left\{ \begin{array}{l} m a = F - F _ {r} - m g \sin \alpha \\ m g \cos \alpha - P = 0 \end{array} \right. \Rightarrow$$ $$a = \frac {F - F _ {r} - m g \sin \alpha}{m} = 6. 2 \frac {m}{s ^ {2}}$$

Answer: $a = 6.2 \frac{m}{s^2}$