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Nuetrik [128]
3 years ago
10

A 13.6 kg block is tied at the top of an incline to a tree. If the incline is 35.5 degrees and the coefficient of friction betwe

en the sled and the incline is .45, What is the tension force between the block and the tree
Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer:

Explanation:

ASSUMING that block = sled AND that the rope is parallel to the slope.

The force acting parallel due to the weight is

13.6(9.81)sin35.5 = 77.475 N

The maximum friction force is

(0.45)13.6(9.81)cos35.5 = 48.877 N

If rope tension is T

77.475 - 48.877 < T < 77.475 + 48.877

            28.6 N < T < 126 N

28.6 N will occur if the block is on the verge of sliding downhill

126 N will occur if the block is on the verge of sliding uphill

Could be any value between them.

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\displaystyle \vec{d}=

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<u>Displacement Vector</u>

Suppose an object is located at a position  

\displaystyle P_1(x_1,y_1)

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\displaystyle \vec{d}=

If the position is given as magnitude-angle data ( z , α), we can compute its rectangular components as

\displaystyle x=z\ cos\alpha

\displaystyle y=z\ sin\alpha

The question describes the situation where the initial point is the base of the mountain, where both components are zero

\displaystyle P_1(0,0)

The final point is given as a 520 m distance and a 32-degree angle, so  

\displaystyle x_2=520\ cos32^o= 440.99\ m

\displaystyle y_2=520\ sin32^o=275.6\ m

The displacement is

\displaystyle \vec{d}=

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3 years ago
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It's velocity when it strikes the ground is. D. 232.9 kg.m/s<span>.
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1. You released a pendulum of mass 1kg from a height of 0.05m
photoshop1234 [79]

a. The speed of the pendulum when it reaches the bottom is 0.9 m/s.

b. The height reached by the pendulum is 0.038 m.

c. When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

<h3>Kinetic energy of the pendulum when it reaches bottom</h3>

K.E = 100%P.E - 18%P.E

where;

  • P.E is potential; energy

K.E(bottom) = 0.82P.E

K.E(bottom) = 0.82(mgh)

K.E(bottom) = 0.82(1 x 9.8 x 0.05) = 0.402 J

<h3>Speed of the pendulum</h3>

K.E = ¹/₂mv²

2K.E = mv²

v² = (2K.E)/m

v² = (2 x 0.402)/1

v² = 0.804

v = √0.804

v = 0.9 m/s

<h3>Final potential energy </h3>

P.E = 100%K.E - 7%K.E

P.E = 93%K.E

P.E = 0.93(0.402 J)

P.E = 0.374 J

<h3>Height reached by the pendulum</h3>

P.E = mgh

h = P.E/mg

h = (0.374)/(1 x 9.8)

h = 0.038 m

<h3>when the pendulum stops</h3>

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Thus, the speed of the pendulum when it reaches the bottom is 0.9 m/s.

The height reached by the pendulum is 0.038 m.

When the pendulum no longer swing at all, all the kinetic energy of the pendulum has been used to overcome frictional force.

Learn more about pendulum here: brainly.com/question/26449711
#SPJ1

5 0
2 years ago
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