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bekas [8.4K]
3 years ago
15

Two blocks of mass m1 = 8.0 kg and m3 = 6.3 kg, connected by a rod of mass m2 = 1.8 kg, are sitting on a very low-friction surfa

ce, and you push to the left on the right block (mass m1) with a constant force of (-30 , 0, 0)N. See figure below. (As usual, +x is to the right and +y is up. Ignore effects due to friction between the objects)
What is the force F3 exerted by the rod on the block of mass m3?
Physics
1 answer:
Tom [10]3 years ago
5 0
Since the three bodies are connected properly and the force is exerted at the end of m1, the force is channeled to the other bodies without any losses. If the force on m1 is 30 N, then the force F3 exerted by the rod on the block of mass m3 is also 30 N.
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A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t
Vanyuwa [196]

Answer:

 x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm  = 0. 1 m

now,

we know

m g = k x

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0.4 x 9.8 = k x 0.1

  k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

W = \dfrac{1}{2}kx'^2

20= \dfrac{1}{2}\times 39.2\times x'^2

 x'² = 1.0204

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hence, the spring is stretched to 1.01 m from the second mass.

 

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3 years ago
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Answer:

Force to stretch the wire is 250 N

Explanation:

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\frac{F L}{\Delta L A} = constant

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\frac{F_1 L}{\Delta L_1 A} = \frac{F_2 L}{\Delta L_2 A}

so we can write it as

F_2 = \frac{F_1 L_2}{L_1}

F_2 = \frac{50 (0.05)}{0.01}

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Answer:

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