Respuesta :
Let m1=10kg and m2=5kg and for our calculations assume right is positive and up is positive (note: for block hanging, the x axis is vertical so tilt your head to help)
For m1
Sigma Fx = ma
T - m1gsin35 = m1a where T = tension
For m2
m2g - T = m2a
Add equation together
m1a + m2a = T-m1gsin35 + m2g - T
a(m1 + m2) = m2g - m1gsin35
a= (5*9.8 - 10*9.8*sin35)/(10 + 5)
a= -0.48m/s/s
So the system is moving in the opposite direction of our set coordinate system where we said right positive, its negative so its moving left therefore down the ramp
The acceleration of the [tex]10\,{\text{kg}}[/tex] block is [tex]\boxed{ - 1.738\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex] . The negative sign shows that the block in accelerating in downward direction.
Further Explanation:
The [tex]5\,{\text{kg}}[/tex] block is hanging from the pulley whereas the [tex]10\,{\text{kg}}[/tex] block is kept on the inclined plane. Consider that the [tex]5\,{\text{kg}}[/tex] bock moves downward with acceleration as shown in the free body diagram attached below:
The force balancing equation for [tex]5\,{\text{kg}}[/tex] block is given as:
[tex]\begin{aligned}{F_{net}}&=m\times a \hfill\\{m_1}g-T&={m_1}a\hfill\\\end{aligned}[/tex]
Here, [tex]{m_1}[/tex] is the mass of the block, [tex]T[/tex] is the tension developed in the string and [tex]a[/tex] is the acceleration of the block.
Rearrange the above expression for [tex]T[/tex].
[tex]T={m_1}\left({g-a}\right)[/tex] …… (1)
Now, the force balancing equation for the [tex]10\,{\text{kg}}[/tex] block kept on the inclined plane.
[tex]T-{m_2}g\sin\theta={m_2}a[/tex]
Substitute [tex]{m_1}\left({g-a}\right)[/tex] for [tex]T[/tex] in above expression:
[tex]\begin{aligned}{m_1}\left({g-a}\right)-{m_2}g\sin\theta&={m_2}a\hfill\\\left({{m_1}+{m_2}}\right)a&={m_1}g-{m_2}g\sin\theta\hfill\\a&=\frac{{{m_1}g-{m_2}g\sin\theta }}{{\left( {{m_1}+{m_2}}\right)}}\hfill\\\end{aligned}[/tex]
Substitute [tex]5\,{\text{kg}}[/tex] for [tex]{m_1}[/tex], [tex]10\,{\text{kg}}[/tex] for [tex]{m_2}[/tex], [tex]9.8\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex] for [tex]g[/tex] and [tex]50^\circ[/tex] for [tex]\theta[/tex] in above expression.
[tex]\begin{aligned}a&=\frac{{5\times 9.8-10\times9.8\times\sin50}}{{\left({5+10}\right)}}\\&=\frac{{49-75.07}}{{15}}\\&=- 1.738\,{{\text{m}}\mathord{\left/ {\vphantom {{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right. \kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}[/tex]
Thus, the acceleration of the [tex]10\,{\text{kg}}[/tex] block is [tex]\boxed{ - 1.738\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}[/tex] . The negative sign shows that the block in accelerating in downward direction.
Learn More:
1. A mug rests on an inclined surface, as shown in (figure 1), θ=13 https://brainly.com/question/5948697
2. A 1500-kg car accelerates from rest to 25 m/s in 7.0 s. What is the average power delivered by the engine? (1 hp = 746 w) https://brainly.com/question/7956557
3. Why is it important to define a frame of reference https://brainly.com/question/526888
Answer Details:
Grade: College
Subject: Physics
Chapter: Newton’s law of Motion
Keywords:
Frictionless ramp, 10kg block slide, released from rest, acceleration, positive value, up the rump, negative value, 5kg block, force balancing.
