Question

the second question is also wrong. I should cancel this subscription. please answer it again with different person

A block of mass m1= 4.00 kg is released from rest on an inclined plane that makes an angle theta= 30 degrees above the horizontal. The coefficient of kinetic friction between the block and the plane is Uk= 0.24. The block is attached to a second block of mass m2= 5.00 kg that hangs freely by a string that passes over a frictionless, massless pulley.

a. Use Newton's 2nd law to write the equations of motion for each mass. To recieve full credit, you must draw a free-body diagram for each mass. Be sure to include a coordinate system and clearly label all forces for each diagram

b. Determine the acceleration of each mass and the tension in the string.

Please show your work clearly and write step by step solution including numeric substitutions after variable manipulation and etc. To make it easier, please do this in paper and include the pictures. thanks

Answer 1

2 2 2

on m1

$\sum\mathrm{F_{x}}: \begin{matrix} & \end{matrix} \mathrm{T_{1}-f_{k,1}-W_{1}\sin\theta=m_{1}a_{1}} \begin{matrix} & ; && \end{matrix} \mathrm{f_{k,1}=\mu N_{1}}$

$\sum\mathrm{F_{y}}: \begin{matrix} & \end{matrix} \mathrm{N_{1}-W_{1}\cos\theta=0} \begin{matrix} & --> && \end{matrix} \mathrm{N_{1}=W_{1}\cos\theta}$

then

$\mathrm{T_{1}-\mu W_{1}\cos\theta-W_{1}\sin\theta=m_{1}a_{1}} \begin{matrix} & --> && \end{matrix} \mathrm{T_{1}-W_{1}\left( \mu \cos\theta+\sin\theta \right)=m_{1}a_{1}}$

on m2

$\sum\mathrm{F_{x}=0N}$

$\sum\mathrm{F_{y}}: \begin{matrix} & \end{matrix} \mathrm{T_{2}-W_{2}=-m_{2}a_{2}}$

so,

$\left\{\begin{matrix} \begin{matrix} \mathrm{T_{1}-W_{1}\left( \mu \cos\theta+\sin\theta \right)=m_{1}a_{1}} & (1) \end{matrix} \\ \begin{matrix} \mathrm{T_{2}-W_{2}=-m_{2}a_{2}} &&&&&&&& (2) \end{matrix} \end{matrix}\right.$

if we consider

$\mathrm{T_{1}=T_{2}=T} \begin{matrix} && \mathrm{and} && \end{matrix} \mathrm{a_{1}=a_{2}=a}$

then,

$\left\{\begin{matrix} \begin{matrix} \mathrm{T-W_{1}\left( \mu \cos\theta+\sin\theta \right)=m_{1}a} & (1) \end{matrix} \\ \begin{matrix} \mathrm{-T+W_{2}=m_{2}a} &&&&&&&& (2) \end{matrix} \end{matrix}\right.$

combining both equations

$\mathrm{a=\frac{W_{2}-W_{1}\left( \mu\cos\theta+\sin\theta \right)}{m_{1}+m_{2}}}$

where,

• $\mathrm{m_{1}=4.00Kg} \begin{matrix} && \mathrm{and} &&\end{matrix} \mathrm{m_{1}=5.00Kg}$
• $\mathrm{W_{1}=m_{1}g=\left( 4.00Kg \right)\left( 9.81\tfrac{m}{s^{2}} \right)=39.24\tfrac{m}{s^{2}}}$
• $\mathrm{W_{2}=m_{2}g=\left( 5.00Kg \right)\left( 9.81\tfrac{m}{s^{2}} \right)=49.05\tfrac{m}{s^{2}}}$
• $\mathrm{\theta=30^{\circ}} \begin{matrix} & --> &&\end{matrix} \sin\left( 30^{\circ} \right)=0.50 \begin{matrix} && \mathrm{and} &&\end{matrix} \cos\left( 30^{\circ} \right)=0.87$

thus, acceleration of blocks

$\mathrm{a=\frac{49.05N-\left( 39.24N \right)\left[ (0.24)(0.87)-0.50 \right]}{4.00Kg+5.00Kg}=2.36\tfrac{m}{s^{2}}}$

and tension on string

$\mathrm{T_{2}-W_{2}=-m_{2}a_{2}}$

$\mathrm{T_{2}=W_{2}-m_{2}a_{2}=49.05N-\left( 5.00Kg \right)\left( 2.36\tfrac{m}{s^{2}} \right)=37.25N}$