Question

learn more remarks the acceleration of the second block is the same as that of the first, but negative. when m₂ gets very large compared with m₁, the acceleration of the system approaches g, as expected, because m₂ is falling nearly freely under the influence of gravity. indeed, m₂ is only slightly restrained by the much lighter m₁. question suppose you weigh 760 n (about 170 lbs) but would find it hard to lift an object weighing 500 n (about 115 lbs). consider how you could use this simple machine to raise the object, and what force you would have to exert to do it. (a) with the pulley hooked to the ceiling, you could pull down on the rope, as if pulling yourself up along the rope. to raise the object tied to the end of the rope you would need to exert a force of n downward. (b) with the pulley attached to the object and one end of the rope attached to the ceiling, you could lift the object by pulling with a force of n directly upward on your end of the rope. exercise hints: getting started | im stuck! suppose that in the same atwood setup another string is attached to the bottom of m₁ and a constant force f is applied, retarding the upward motion of m₁. if m₁ = 5.90 kg and m₂ = 11.80 kg, what value of f will reduce the acceleration of the system by 50%? n downward resources read it

Explanation:

Step1: Analyze pulley - basic principle

For a single - fixed pulley, the force required to lift an object is equal to the weight of the object.

Step2: Analyze movable pulley principle

For a single movable pulley, the force required to lift an object is half of the weight of the object.

Step3: Calculate force for single - fixed pulley (part a)

The weight of the object is $W = 500\ N$. For a single - fixed pulley, the force $F_1$ required to lift the object is equal to the weight of the object. So $F_1=500\ N$.

Step4: Calculate force for single - movable pulley (part b)

For a single movable pulley, the force $F_2$ required to lift the object is $F_2=\frac{W}{2}$. Substituting $W = 500\ N$, we get $F_2 = 250\ N$.

Step5: Calculate acceleration of Atwood's machine without additional force

The acceleration of an Atwood's machine is given by $a=\frac{m_2 - m_1}{m_1 + m_2}g$, where $g = 9.8\ m/s^2$, $m_1=5.90\ kg$ and $m_2 = 11.80\ kg$.
\[a=\frac{11.80 - 5.90}{5.90+11.80}\times9.8=\frac{5.90}{17.70}\times9.8=\frac{1}{3}\times9.8\ m/s^2\]

Step6: Calculate the new acceleration

We want to reduce the acceleration by 50%, so the new acceleration $a'=\frac{a}{2}=\frac{1}{6}\times9.8\ m/s^2$.

Step7: Apply Newton's second law with additional force

Let the additional force be $f$. According to Newton's second law for the Atwood's machine with the additional force: $m_2g-(m_1g + f)=(m_1 + m_2)a'$.
\[11.80\times9.8-(5.90\times9.8 + f)=(5.90 + 11.80)\times\frac{1}{6}\times9.8\]
\[11.80\times9.8-5.90\times9.8 - f=(5.90 + 11.80)\times\frac{1}{6}\times9.8\]
\[5.90\times9.8 - f = 17.70\times\frac{1}{6}\times9.8\]
\[f=5.90\times9.8-17.70\times\frac{1}{6}\times9.8\]
\[f = 57.82-28.91=28.91\ N\]

Answer:

(a) 500
(b) 250
(c) 28.91