Question

If the blocks are released from rest, which way does the 10 kg block slide, and what is its acceleration? enter a positive value if 10 kg block accelerates up the rump, and a negative value otherwise. express your answer to two significant figures and include the appropriate units.

Answer 1

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

Answer 2

The acceleration of the $10\,{\text{kg}}$ block is $\boxed{ - 1.738\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ . The negative sign shows that the block in accelerating in downward direction.

Further Explanation:

The $5\,{\text{kg}}$ block is hanging from the pulley whereas the $10\,{\text{kg}}$ block is kept on the inclined plane. Consider that the $5\,{\text{kg}}$ bock moves downward with acceleration   as shown in the free body diagram attached below:

The force balancing equation for $5\,{\text{kg}}$ block is given as:

$\begin{aligned}{F_{net}}&=m\times a \hfill\\{m_1}g-T&={m_1}a\hfill\\\end{aligned}$

Here, ${m_1}$ is the mass of the block, $T$ is the tension developed in the string and $a$ is the acceleration of the block.

Rearrange the above expression for $T$.

$T={m_1}\left({g-a}\right)$                              …… (1)

Now, the force balancing equation for the $10\,{\text{kg}}$ block kept on the inclined plane.

$T-{m_2}g\sin\theta={m_2}a$

Substitute ${m_1}\left({g-a}\right)$ for $T$ in above expression:

$\begin{aligned}{m_1}\left({g-a}\right)-{m_2}g\sin\theta&={m_2}a\hfill\\\left({{m_1}+{m_2}}\right)a&={m_1}g-{m_2}g\sin\theta\hfill\\a&=\frac{{{m_1}g-{m_2}g\sin\theta }}{{\left( {{m_1}+{m_2}}\right)}}\hfill\\\end{aligned}$

Substitute $5\,{\text{kg}}$ for ${m_1}$, $10\,{\text{kg}}$ for ${m_2}$, $9.8\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}$ for $g$ and $50^\circ$ for $\theta$ in above expression.

$\begin{aligned}a&=\frac{{5\times 9.8-10\times9.8\times\sin50}}{{\left({5+10}\right)}}\\&=\frac{{49-75.07}}{{15}}\\&=- 1.738\,{{\text{m}}\mathord{\left/ {\vphantom {{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right. \kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{aligned}$

Thus, the acceleration of the $10\,{\text{kg}}$ block is $\boxed{ - 1.738\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}}$ . 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 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) brainly.com/question/7956557

3. Why is it important to define a frame of reference 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.