Answer to Question #157318 in Mechanics | Relativity for Manisha Nayak

Question

A crate of mass 40.0 kg is pulled by a force of 2000 N, up an inclined plane which 

makes an angle of 30º with the horizon. The coefficient of kinetic friction between the 

plane and the crate is miu.k = 0.20. If the crates starts from rest, calculate its speed after it 

has been pulled 10.0 m. Draw the free body diagram. Take 

g=10.0 ms-² .

Answer 1

Let's draw a FBD:

Let's apply Newton's Second Law of Motion in projections on axis $x$- and $y$:

$F_{pull}-mgsin\theta-F_{fr}=ma,$ $N=mgcos\theta,$ $F_{pull}-mgsin\theta-\mu_k mgcos\theta=ma.$

From this equation, we can find the acceleration of the crate:

$a=\dfrac{F_{pull}-mg(sin\theta+\mu_kcos\theta)}{m},$ $a=\dfrac{2000\ N-40\ kg\cdot 10\ \dfrac{m}{s^2}\cdot(sin30^{\circ}+0.2\cdot cos30^{\circ})}{40\ kg}=43.3\ \dfrac{m}{s^2}.$

Finally, we can find the speed of the crate after it has been pulled 10.0 m:

$v_f^2=v_i^2+2ad,$ $v_f=\sqrt{2ad}=\sqrt{2\cdot 43.3\ \dfrac{m}{s^2}\cdot10.0\ m}=29.43\ \dfrac{m}{s}.$

Answer:

$v_f=29.43\ \dfrac{m}{s}.$