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SOVA2 [1]
3 years ago
15

A light rope is attached to a block with mass 3.60 kg that rests on a frictionless, horizontal surface. The horizontal rope pass

es over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 18.8 N .
(a) Draw two free-body diagrams: one for each block.
(b) What is the acceleration of either block?
(c) Find m.
(d) How does the tension compare to the weight of the hanging block?
Physics
1 answer:
Readme [11.4K]3 years ago
4 0

Answer and Explanation:

(a) The fre-body diagrams for each block is shown below. In the block of mass 3.60 kg, there are 3 forces acting on it: horizontal force due to the rope (F_{t}), vertical gravitational force (F_{g}) and vertical normal force (F_{n}), due to the surface. Since there is no vertical movement, F_{g} and F_{n} cancels it out. So, for this block, net force is horizontal due to the rope F_{t}.

The block of mass m is hanging from the pulley, so there is the force of the rope (F_{t}) and the gravitational force (F_{g}). Both are vertical, because there is no surface "holding" block m.

(b) Since both blocks are attached to each other, the acceleration will be the same. To calculate it, we use the Second Law of Motion:

F_{r}=m.a

a=\frac{F_{r}}{m}

a=\frac{18.8}{3.6}

a = 5.22

The acceleration of either block is 5.22 m/s².

(c) Block m has 2 forces acting on it: tension and gravitational force. Gravitational force is the force of attraction the Earth does over an object. It is calculated as the product of mass and gravitational acceleration, which has magnitude g = 9.8 m/s².

Suppose positive referential is going up. To determine mass:

F_{r}=m.a

F_{t}-F_{g}=m.a

F_{t}-m.g=m.a

18.8-9.8m=5.22m

15.02m=18.8

m = 1.25

Block m has 1.25 kg.

(d) Gravitational force is also called weight. So, as described above: F_{g}=m.g.

The weight for the hanging block is

F_{g}=1.25*9.8

F_{g}= 12.25 N

Comparing tension and weight:

\frac{12.25}{18.8} ≈ 0.65

We can see that, weight of the hanging block is almost 0.65 times smaller than the tension on the rope.

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Answer:

The answer is given below

Explanation:

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v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\  Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s

The final velocity of cart 2 after collision is given as:

v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\  Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s

b) Using the law of conservation of energy:

\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm

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