Answer to Question #140527 in Mechanics | Relativity for Tanvir Nijjar

Question

A dynamics cart sits on a lab desk has a mass of 900 g. It is connected to a suspended 500 g mass by a string that is threaded through a frictionless pulley. The coefficient of friction between the lab desk and the dynamics cart is 0.25.
a) Determine the acceleration of the dynamics cart when the 500 g mass is released (Include FBD’s).
b) If the dynamics cart takes 1.3 s to reach the pulley, what distance does it travel?

Answer 1

A) According to Newton's second law for the first body $a\times m_1=T-F_f$ , where $a$ is acceleration, $m$ mass, $T$ is the tension force of the filament, $F_f=m_1\times g\times\mu$ is the friction force, for the second body the equation will be: $-a\times m_2=T-m_2\times g$ ;

$T = a\times m_1+F_f$ , therefore

$-a\times m_2= a\times m_1 +m_1\times g\times \mu-m_2\times g$ ;

$a=-\frac{ m_1\times g\times \mu -m_2\times g}{m_1 m_2}=-\frac{0.9\times 10 \times 0.25-0.5\times 10}{0.9 0.5}\approx 1.96$ m/s$^2$ .

B) Since the distance travelled without the initial speed is $S =\frac {a\times t ^ 2} {2}, S =\frac {1 .96\times 1.3 ^ 2} {2 }\approx 1.66$ m.