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Vikentia [17]
3 years ago
12

A pile driver lifts a 450 kg weight and then lets it fall onto the end of a steel pipe that needs to be driven into the ground.

A fall of 1.5 m drives the pipe in 45 cm.
What is the average force exerted on the pipe?
Physics
2 answers:
nikitadnepr [17]3 years ago
6 0
The gravitational potential energy becomes work driving the pipe 

<span>m g h = f d </span>

<span>450 * g * 1.5 = f * .45</span>
Vika [28.1K]3 years ago
6 0

Answer:

F = 14700N

Explanation:

Hello! Let's solve this!

We have the following data:

vi = 0m / s

dy = 1.5m

m = 450kg

dx = 0.45m

First we need to know the speed of the impeller.

We can use the following formula

vf^{2}=vi^{2}+2*g*dy

We get the vf

vf=\sqrt{2*g*dy}

vf=\sqrt{2*9.8m/s^{2}*1.5m }

vf = 4.42m / s

Then the kinetic energy is equal to the work done.

KE=\frac{1}{2}*v^{2} *m

KE=1/2*5.42m/s^{2}*450kg

We cleared KE

KE = W = 6615J

From the working formula we will clear the force exerted.

W = F * dx

F = W / dx = 6615J / 0.45m

F = 14700N

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Simora [160]

Answer:

a)  W = 5360 J  b)  μ = 0.29

Explanation:

a) The work is, bold indicate vectors

         W = F. d ​​= F d cos T

         W = 40 268 cos 60

         W = 5360 J

b) The force of friction and opposes the movement of the body and if the speed of the body is constant implies that the external force in the direction of the movement is equal to the force of friction, or which work is the same

         W_{fr} = -5360J

The negative sign is because the force of friction is contrary to the direction of movement

c) Let's write the work of the friction force

           W_{fr} = fr d cos 180

           fr = μ N

Since the body is on a horizontal surface, from Newton's second law on the Y axis

          N-W = 0

          N = W

          fr = μ W

         W_{fr} = - μ W d

         μ = - W_{fr} / W d

         μ = - (-5360) / 69 268

         μ = 0.29

7 0
3 years ago
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Answer:

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Explanation:

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Answer:

2.8s

Explanation:

=>a = v-u/t

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