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Inga [223]
3 years ago
11

Dwight uses a spring (k = 70 N/m) on a horizontal surface to make a launcher for model cars. The spring is attached to a holder

that has a mass of 0.3 kg. If he compresses the spring and releases it, the launcher will move back and forth in periodic motion.
a. If he compresses the spring by 0.4 m and releases it, what is the acceleration of the launcher as the spring reaches its natural length?

b. What is the maximum kinetic energy fo the launcher?

c. What is the maximum speed of the launcher?
Physics
2 answers:
kramer3 years ago
8 0

Part A)

As we know that spring force is given by

F = kx

here x = stretch in the spring from natural length

So here  when spring reaches to its natural length

Force due to spring = 0

so acceleration = 0

Part b)

When spring is compressed from its natural length it will have elastic potential energy in it

so it is given by

U = \frac{1}{2}kx^2

now we know that there is no friction in it so maximum kinetic energy of the launcher must be equal to the elastic potential energy of the spring

KE = \frac{1}{2}kx^2

here we have

k = 70 N/m

x = 0.4 m

KE = \frac{1}{2}(70)(0.4)^2

KE = 5.6 J

Part c)

Now to find the speed we know that

KE = \frac{1}{2} mv^2

5.6 = \frac{1}{2}0.3v^2

v = 6.11 m/s

so its speed is 6.11 m/s

lubasha [3.4K]3 years ago
4 0

a) In the very beginning its acceleration is quite big, but when the spring is stretched to the normal length, its acceleration is 0.

b) Speed will be

KE=1/2 mv^2=1/2 kx^2=5.6 J

c) v=\sqrt{\frac{2KE}{m} }=6.11 m/s

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umka21 [38]

Answer:

6.58m

Explanation:

The kinetic energy = Workdone on the roller

Workdone = Force * distance

Given

KE = Workdone = 362J

Force = 55N

Required

Distance

Substitute into the formula;

Workdone = Force * distance

362 = 55d

d = 362/55

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Hence the student must push at a distance of 6.58m

3 0
3 years ago
At t=0, a block A of mass 8 kg and block B of mass 16 kg are both at position x=0 . Block A is at rest, and block B is moving at
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The center of mass of the two objects is the average position of the parts of the two object system

The center of mass of block <em>A</em>, and block <em>B</em>  after displacement of block <em>B</em> is at <u>20 m from block </u><u><em>A</em></u>

<em />

Reason:

The given parameters are;

The position of block A and block B at t = 0 is x = 0

The mass of block A, m₁ = 8 kg

Mass of block B, m₂ = 16 kg

Speed of block <em>A</em> = 0 m/s

Speed of block <em>B</em>, v₂ = 10 m/s

Location of the center of mass of the two object at t = 3 s; Required

Solution;

The location of block <em>A</em>, after 3 s is x₁ = 0 (block A is at rest)

The location of block <em>B</em>, = v₂ × t

The location of block <em>B</em>, after 3 s is x₂ = 10 m/s × 3 s = 30 m

The center of mass of two masses are given as follows;

x_{cm} = \dfrac{m_1 \cdot x_1 +m_2\cdot x_2}{m_1 + m_2}

x_{cm} = \dfrac{8  \times0 + 16 \times  30}{8 + 16} = 20

The center of mass of the two objects is at at the position x = <u>20 m</u> (from block <em>A</em>)

Learn more about the center of mass here:

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