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2 years ago

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A 2. 5 kg block is launched along the ground by a spring with a spring constant of 56 N/m. The spring is initially compressed 0.

75 m. Disregarding friction, how fast will the block move after the spring is released all the way and the block slides away from it? 3. 5 m/s 4. 1 m/s 13 m/s 16 m/s.

2 answers:

Gnoma [55]2 years ago

7 0

Answer:

A- 3.5

Explanation: Edge

OLga [1]2 years ago

3 0

The speed of the spring when it is released is 3.5 m/s.

The given parameters:

<em>Mass of the block, m = 2.5 kg</em>
<em>Spring constant, k = 56 N/m</em>
<em>Extension of the spring, x = 0.75 m</em>

The speed of the spring when it is released is calculated by applying the principle of conservation of energy as follows;

$K.E = U_x\\\\\frac{1}{2} mv^2 = \frac{1}{2} kx^2\\\\mv^2 = kx^2\\\\v^2 = \frac{kx^2}{m} \\\\v = \sqrt{\frac{kx^2}{m} } \\\\v = \sqrt{\frac{56 \times 0.75^2}{2.5} } \\\\v = 3.5 \ m/s$

Thus, the speed of the spring when it is released is 3.5 m/s.

Learn more about conservation of energy here: brainly.com/question/166559

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A 0.01 kg spring toy is compressed 0.02 m and released vertically. The toy is measured to reach 0.25 m in the air. Determine the

Scorpion4ik [409]

Answer:

122.5 N/m

Explanation:

According to the law of conservation of energy, if there is no air resistance or frictional forces, the initial elastic potential energy of the spring toy is entirely converted into gravitational potential energy when the toy reaches the highest point.

Therefore, we can write:

$\frac{1}{2}kx^2=mgh$

where the term on the left is the initial elastic potential energy while the term on the right is the gravitational potential energy, and where

k is the spring constant

x = 0.02 m is the compression of the spring

m = 0.01 kg is the mass of the toy

h = 0.25 m is the height reached by the toy

$g=9.8 m/s^2$ is the acceleration due to gravity

Solving for k,

$k=\frac{2mgh}{x^2}=\frac{2(0.01)(9.8)(0.25)}{(0.02)^2}=122.5 N/m$

8 0

2 years ago

A 51-kg woman runs up a vertical flight of stairs in 5.0 s. Her net upward displacement is 5.0 m. Approximately, what average po

Vika [28.1K]

Answer:

The average power the woman exerts is 0.5 kW

Explanation:

We note that power, P = The rate at which work is done = Work/Time

Work = Energy

The total work done is the potential energy gained which is the energy due to vertical displacement

Given that the vertical displacement = 5.0 m, we have

Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h

m = 51 kg

g = Constant = 9.81 m/s²

h = 5.0 m

Also, time, t = 5.0 s

Total work done = 51 kg × 9.81 m/s²× 5 m = 2501.55 kg·m²/s² = 2501.55 J

P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.

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3 years ago

The roller coaster car has a mass of 700 kg, including its passenger. If it is released from rest at the top of the hill A, dete

sweet [91]

Answer:

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

$F_n = 1.72 \times 10^4$

Explanation:

Since we need to cross both the loops so least speed at the bottom must be

$v = \sqrt{5 R g}$

also by energy conservation this is gained by initial potential energy

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

so we will have

$\sqrt{2gh} = \sqrt{5Rg}$

now we have

$h = \frac{5R}{2}$

here we have

R = 7.5 m

so we have

$h = \frac{5(7.5)}{2}$

$h = 18.75 m$

Now when it will reach at point B then its normal force is just equal to ZERO

$N_B = 0$

now when it reach point C then the speed will be

$mgh - mg(2R_c) = \frac{1}{2]mv_c^2$

$v_c^2 = 2g(h - 2R_c)$

$v_c = 13.1 m/s$

now normal force at point C is given as

$F_n = \frac{mv_c^2}{R_c} - mg$

$F_n = \frac{700\times 13.1^2}{5} - (700 \times 9.8)$

$F_n = 1.72 \times 10^4$

7 0

3 years ago

A torsion-bar spring consists of a prismatic bar, usually of round cross section, that is twisted at one end and held fast at th

ra1l [238]

Answer:

d₁ = 0.29 in

d₂ = 0.505 in

Explanation:

Given:

T = 1500 lbf in

L = 10 in

x = 0.5 L = 5 in

$T_{1} =\frac{T(L-x)}{L} =\frac{1500*(10-5)}{10} =750lbfin$

First case: T = T₁ + T₂

T₂ = T - T₁ = 1500 - 750 = 750 lbf in

If the shafts are in series:

θ = θ₁ + θ₂

θ = ((T₁ * L₁)/GJ) + ((T₂ * L₂)/GJ)

Second case: If d₁ ≠ d₂

θ = ((T₁ * L₁)/GJ₁) + ((T₂ * L₂)/GJ₂) = 0 (eq. 1)

t₁ = t₂

$\frac{16T_{1} }{\pi d_{1}^{3} } =\frac{16T_{2} }{\pi d_{2}^{3} }$ (eq. 2)

T₁ + T₂ = 1500 (eq. 3)

θ₁ first case = θ₁ second case

Replacing:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{1}*3.7 }{G(\frac{\pi }{32})*d_{1} ^{4} }\\T_{1} =16216d_{1} ^{4}$

The same way to θ₂:

$\frac{750*5}{G(\frac{\pi }{32})*0.5^{4} } =\frac{T_{2}*6.3 }{G(\frac{\pi }{32})*d_{2} ^{4} } \\T_{2} =9523.8d_{2} ^{4}$

From equation 2, we have:

d₁ = 0.587 * d₂

From equation 3, we have:

d₂ = 0.505 in

d₁ = 0.29 in

7 0

3 years ago

A 2.00 kg ball is thrown upward at Some unknown angle from the top of a 20.0 M high building If the initial magnitude of the vel

mixer [17]

Answer:

792 J

Explanation:

The total energy of the ball is E = U + K where U = potential energy = mgh and K = kinetic energy = 1/2mv²

E = mgh + 1/2mv² where m = mass of ball = 2.0 kg, g = acceleration due to gravity = 9.8 m/s², h = height of building = 20.0 m, v = initial velocity of ball = 20.0 m/s.

So, substituting the values of the variables into E, we have

E = mgh + 1/2mv²

= 2.00 kg × 9.8 m/s² × 20.0 m + 1/2 × 2.00 kg × (20.0 m/s)²

= 392 J + 400 J

= 792 J

6 0

3 years ago