Question

Consider the two-body situation at the right. A 3.50x103-kg crate (m1) rests on an inclined plane and is connected by a cable to a 1.00x103-kg mass (m2). This second mass (m2) is suspended over a pulley. The incline angle is 30.0° and the surface has a coefficient of friction of 0.210. Determine the acceleration of the system and the tension in the cable.

Answer 1

Answer:$a=2.42 m/s^2$

Explanation:

Given

mass $m_1=3.50\times 10^{3} kg$

$m_2=1.00\times 10^{3} kg$

$\theta$(inclination)=$30^{\circ}$

$\mu =0.21$

Let T be the tension in the rope

From Diagram

$m_1g\sin \theta -T-f_r=m_1a$-----------------1

where $f_r=friction\ force$

$f_r=\mu m_g\cos \theta$

For block $m_2$

$T=m_2a$-----------2

From 1 & 2

$m_1g\sin \theta -m_2a-\mu m_1g\cos \theta =m_1a$

$m_1g(\sin \theta -\mu \cos \theta )=(m_1+m_2)a$

$\frac{9.8\times 3.5\times 10^3}{4.5\times 10^3}(0.5-\0.21\cos 30)=4.5a$

$a=2.42 m/s^2$