When a 40-N force, parallel to the incline and directed up the incline, is applied to a crate on a frictionless incline that is 30° above the horizontal, the acceleration of the crate is 2.0 m/s^2, up the incline. The mass of the crate is ?

Apr 20, 2018

$m \cong 5.8 k g$

Explanation:

The net force up the incline is given by

${F}_{\text{net}} = m \cdot a$

${F}_{\text{net}}$ is the sum of the 40 N force up the incline and the component of the object's weight, $m \cdot g$, down the incline.

${F}_{\text{net}} = 40 N - m \cdot g \cdot \sin 30 = m \cdot 2 \frac{m}{s} ^ 2$

Solving for m,

$m \cdot 2 \frac{m}{s} ^ 2 + m \cdot 9.8 \frac{m}{s} ^ 2 \cdot \sin 30 = 40 N$

$m \cdot \left(2 \frac{m}{s} ^ 2 + 9.8 \frac{m}{s} ^ 2 \cdot \sin 30\right) = 40 N$

$m \cdot \left(6.9 \frac{m}{s} ^ 2\right) = 40 N$

$m = \frac{40 N}{6.9 \frac{m}{s} ^ 2}$

Note: the Newton is equivalent to $k g \cdot \frac{m}{s} ^ 2$. (Refer to F=ma to confirm this.)

$m = \frac{40 k g \cdot \cancel{\frac{m}{s} ^ 2}}{4.49 \cancel{\frac{m}{s} ^ 2}} = 5.8 k g$

I hope this helps,
Steve

$5.793 \setminus k g$

Explanation:

Given that a force $F = 40 \setminus N$ is applied on the crate of mass $m$ kg to cause it to move with an acceleration $a = 2 \setminus \setminus {\textrm{\frac{m}{s}}}^{2}$ up the plane inclined at an angle $\setminus \theta = {30}^{\setminus} \circ$ with the horizontal.

Applying Newton's second law , the net force acting on the crate moving up the inclined plane

${F}_{\setminus \textrm{\ne t}} = m a$

$F - m g \setminus \sin \setminus \theta = m a$

$F = m \left(a + g \setminus \sin \setminus \theta\right)$

$m = \setminus \frac{F}{a + g \setminus \sin \setminus \theta}$

$= \setminus \frac{40}{2 + 9.81 \setminus \sin {30}^{\setminus} \circ}$

$= \setminus \frac{40}{6.905}$

$= 5.793 \setminus k g$