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Mamont248 [21]
3 years ago
6

Three sleds (30kg sled connected by tension rope B to 20kg sled connected by tension rope A to 10kg sled) are being pulled horiz

ontally on frictionless horizontal ice using horizontal ropes. The pull is horizontal and of magnitude 143N . Required:a. Find the acceleration of the system. b. Find the tension in rope A. c. Find the tension in rope B.
Physics
1 answer:
SOVA2 [1]3 years ago
8 0

Answer:

a) a = 2.383 m / s², b)   T₂ = 120,617 N , c)   T₃ = 72,957 N

Explanation:

This is an exercise of Newton's second law let's fix a horizontal frame of reference

in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.

a) request the acceleration of the system

we can take the sledges together and write Newton's second law

     T = (m₁ + m₂ + m₃) a

    a = T / (m₁ + m₂ + m₃)

     a = 143 / (10 +20 +30)

     a = 2.383 m / s²

b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg

on the sled of m₁ = 10 kg

          T - T₂ = m₁ a

in this case T₂ is the cable tension

           T₂ = T - m₁ a

            T₂ = 143 - 10 2,383

            T₂ = 120,617 N

c) The cable tension between the masses of 20 and 30 kg

            T₂ - T₃ = m₂ a

             T₃ = T₂ -m₂ a

             T₃ = 120,617 - 20 2,383

             T₃ = 72,957 N

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A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i
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Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

\boxed{\mathfrak{range =  \frac{  u  {}^{2}   \sin 2\theta }{g} }}

u is the velocity during its take off and \theta is the angle at which its thrown

Given that

  • u = 8m/ s
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calculating range using the above formula

= \frac{ {8}^{2} \sin2(40)  }{10}

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=  \frac{64 \times 0.985}{10}

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A 60 kg block slides along the top of a 100 kg block with an acceleration of 2.0 m/s2 when a horizontal force F of 340 N is appl
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Answer:

The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

Explanation:

Suppose we find the coefficient of friction and the acceleration of the 100 kg block during the time that the 60 kg block remains in contact.

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Mass of block = 60 kg

Acceleration = 2.0 m/s²

Mass = 100 kg

Horizontal force = 340 N

Let the frictional force be f.

We need to calculate the frictional force

Using balance equation

F-f=ma

Put the value into the formula

340-f=60\times2.0

f=340-60\times2.0

f=220\ N

We need to calculate the coefficient of friction

Using formula of friction force

f= \mu mg

\mu=\dfrac{f}{mg}

\mu=\dfrac{220}{60\times9.8}

\mu =0.37

We need to calculate the acceleration of the 100 kg block

Using formula of newton's law

F = ma

a=\dfrac{F}{m}

a=\dfrac{220}{100}

a=2.2\ m/s^2

Hence, The coefficient of friction and acceleration are 0.37 and 2.2 m/s²

3 0
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