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3 years ago

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A 100. N block sits on a rough horizontal floor. The coefficient of sliding friction between the block and the floor is 0.250. A

horizontal force of 90.0 N acts on the block for 3.00 seconds. Calculate the velocity of the block after 3.00 seconds if it starts from rest.

1 answer:

zaharov [31]3 years ago

8 0

Using the Impulse Theorem, we have:

$F_{r}.\Delta t=m.\Delta v \\ (F-F_{at}).(3-0)= \frac{P}{g} .(v-0) \\ (90-100.0,25).3= \frac{100v}{10} \\ v= \frac{3.65}{10} \\ \boxed {v=19,5 m/s}$

If you notice any mistake in English, please let me know, because I'm not native.

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

5 0

3 years ago

What is the net force on this object f air = 400n (up) fgrav=600n (down)

n200080 [17]

Fnet=F1+F2 or Fnet=F1-F2
So 400n up - 600n down
Fnet= 400-600= -200N

5 0

3 years ago

Read 2 more answers

A 132 cm wire carries a current of 2.2 A. The wire is formed into a circular coil and placed in a B Field of intensity 1 T. a) F

EastWind [94]

Given Information:

Length of wire = 132 cm = 1.32 m

Magnetic field = B = 1 T

Current = 2.2 A

Required Information:

(a) Torque = τ = ?

(b) Number of turns = N = ?

Answer:

(a) Torque = 0.305 N.m

(b) Number of turns = 1

Explanation:

(a) The current carrying circular loop of wire will experience a torque given by

τ = NIABsin(θ) eq. 1

Where N is the number of turns, I is the current in circular loop, A is the area of circular loop, B is the magnetic field and θ is angle between B and circular loop.

We know that area of circular loop is given by

A = πr²

where radius can be written as

r = L/2πN

So the area becomes

A = π(L/2πN)²

A = πL²/4π²N²

A = L²/4πN²

Substitute A into eq. 1

τ = NI(L²/4πN²)Bsin(θ)

τ = IL²Bsin(θ)/4πN

The maximum toque occurs when θ is 90°

τ = IL²Bsin(90)/4πN

τ = IL²B/4πN

torque will be maximum for N = 1

τ = (2.2*1.32²*1)/4π*1

τ = 0.305 N.m

(b) The required number of turns for maximum torque is

N = IL²B/4πτ

N = 2.2*1.32²*1)/4π*0.305

N = 1 turn

8 0

3 years ago

two bumper cars at an amusement park collide together abd get stuck together. assuming that the system of the two bumper cars is

Fiesta28 [93]

The answer to this problem is M1v1-m2v2=(m1+m2)v

7 0

3 years ago

What is the period of a soundwave whose wavelength is 20.0 m? Use values from the book and show ALL of your work.

Liono4ka [1.6K]

<h3><u>Answer;</u></h3>

Period = 1/17 seconds

<h3><u>Explanation;</u></h3>

Wavelength is related to period by the expression:

<em>speed = wavelength / period </em>

If we are given the speed, then we can easily calculate the period at the wavelength of 20 m.

<em>Given the speed of sound wave as 340 m/s </em>

<em>Period = Wavelength/ speed</em>

<em> = 20 m/340 m/s</em>

<em> </em><u><em>= 1/17 seconds</em></u>

7 0

3 years ago