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mote1985 [20]
3 years ago
10

A protester carries his sign of protest, starting from the origin of an xyz coordinate system, with the xy plane horizontal. He

moves 40 m in the negative direction of the x axis, then 20 m along a perpendicular path to his left, and then 25 m up a water tower.
(a) In unit-vector notation, what is the displacement of the sign from start to end?
(b) The sign then falls to the foot of the tower. What is the magnitude of the displacement of the sign from start to this new end?
Physics
1 answer:
OverLord2011 [107]3 years ago
6 0

Answer:

Part A:

Displacement\ vector= -40\hat i+20\hat j+25\hat k

Part B:

Magnitude of displacement=44.721359 m

Explanation:

Given Data:

40 m in negative direction of the x axis

20 m perpendicular path to his left (considering +ve y direction)

25 m up a tower (Considering +ve z direction)

Required:

  • In unit-vector notation, what is the displacement of the sign from start to end.
  • The magnitude of the displacement of the sign from start to this new end=?

Solution:

Part A:

Displacement= (0-40)\hat i+(0+20)\hat j+(0+25)\hat k\\Displacement= -40\hat i+20\hat j+25\hat k

where:

i,j,k are unit vectors

Part B:

Sign falls to foot of tower so z=0

Displacement= -40\hat i+20\hat j+0\hat k

Magnitude of displacement:

Magnitude\ of\ displacement=\sqrt{(-40)^2+(20)^2+0^2} \\Magnitude\ of\ displacement=44.721359\ m

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<h3>Answer:  130 newtons</h3>

===============================================================

Explanation:

We'll need the acceleration first.

  • The initial speed (let's call that Vi) is 8.0 m/s
  • The final speed (Vf) is 0 m/s since Sam comes to a complete stop at the end.
  • This happens over a duration of t = 4.0 seconds

The acceleration is equal to the change in speed over change in time

a = acceleration

a = (change in speed)/(change in time)

a = (Vf - Vi)/(4 seconds)

a = (0 - 8.0)/4

a = -8/4

a = -2

The acceleration is -2 m/s^2, meaning that Sam slows down by 2 m/s every second. Negative accelerations are often associated with slowing down. The term "deceleration" can be used here.

Here's a further break down of Sam's speeds at the four points of interest

  • At 0 seconds, he's going 8 m/s
  • At the 1 second mark, he's slowing down to 8-2 = 6 m/s
  • At the 2 second mark, he's now at 6-2 = 4 m/s
  • At the 3 second mark, he's at 4-2 = 2 m/s
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Next, we'll apply Newton's Second Law of motion

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We just found the acceleration, and the mass is fairly easy as all we need to do is add Sam's mass with the sled's mass to get 60+5.0 = 65 kg

So the force applied must be:

F = m*a

F = 65*(-2)

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This force is negative to indicate it's pushing against the sled's momentum to slow Sam down.

The magnitude of this force is |F| = |-130| = 130 newtons

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The frictional force between the block and the horizontal surface is determined by applying Newton's law;

∑F = ma

F - Ff = ma

Ff = F - ma

Ff = 4 - 2(1.2)

Ff = 4 - 2.4

Ff = 1.6 N

When the applied force increases to 5 N, the magnitude of the block's acceleration is calculated as follows;

F - Ff = ma

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a = 3.4/2

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Answer:

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From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

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Where m is mass

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Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

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Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

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