Answer to Question #241854 in Mechanical Engineering for Papi Chulo

Question

A mass of 9kg draws a second mass of 4.5 kg up a smooth 30° incline by a light cord passing over a smooth pulley. Find the velocities of the two bodies after they have moved a distance of 3m from rest(Use conservation of energy)

Clearly show the following values:

Direction of movement=??

Friction force=??

E_k=??

E_p=??

E_TR=??

E_TE=??

E_FRICTION=??

Energy Equation=??

V_A=??

V_B=??

Answer 1

Direction of movement:

Mass A moves upwards, mass B moves downwards

Frictional force; this is assumed to be zero since the surface is smooth

E_k=

E_p=

Energy conservation requires that the change in potential energy of the system m₁+m₂ is of same magnitude to change in kinetic energy of the system but of a negative sign,

∆K_in=-∆P_out

The two masses are connected and move with the same speed V,

E_friction=0

Energy equation;

$\frac{1}{2}m_1V^{2}+\frac{1}{2}m_2V^{2}=(m_1g∆h_1+m_2g∆h_2+0)$

$(m_1+m_2)V^{2}=2(m_1gh_2Sin\theta+m_2gh_2)$

$V^{2}=\frac{2(m_1gh_2Sin\theta+m_2gh_2)}{(m_1+m_2)}$

$V=\sqrt\frac{2(m_1gh_2Sin\theta+m_2gh_2)}{(m_1+m_2)}$

$V=\sqrt\frac{2(4.5*9.81*3Sin30°+9*9.81*3)}{(4.5+9)}$

$V=7.0036m/s$

$V_A=7.0036m/s$

$V_B=7.0036m/s$