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laiz [17]
3 years ago
14

A 968 g block is released from rest at height h0 above a vertical spring with spring constant k = 440 N/m and negligible mass. T

he block sticks to the spring and momentarily stops after compressing the spring 22.0 cm. How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of h0? (d) If the block were released from height 6h0 above the spring, what would be the maximum compression of the spring?
Physics
1 answer:
olga nikolaevna [1]3 years ago
7 0

Answer:

a) 10.648 J

b) -10.648 J

c) h₀ = 0.9024 m

d) 0.500 m

Explanation:

The work done by the block on the spring is equal to the work done by the spring on the block

W = (1/2)(k)(x²) = 0.5 (440)(0.22²) = 10.648 J

c) The work done by the block on the spring is the change in potential energy of the block during this fall.

The change in potential energy is

Weight × Total distance fallen through by the block = mg(h₀ + 0.22)

This change in potential energy is now equal to the workdone by the spring on the block

W = mg(h₀ + 0.22)

9.4864 h₀ + 2.087008 = 10.648

h₀ = (8.560992/9.4864)

h₀ = 0.9024 m

d) If the block were released from a height of 6h₀, that is,

H = 6h₀ = (6×0.9024) = 5.415 m above the spring

Work done by the block in falling that height + compression = mg(H + x) = (0.968)(9.8)[5.415 + x)

This work is done on the spring

W = (1/2)(k)(x²)

9.4864[5.415 + x] = 0.5(440)(x²)

51.37 + 9.4864x = 220x²

220x² - 9.4864x - 51.37 = 0

Solving the quadratic equation

x = 0.50 m or - 0.47

The negative answer defies the laws of physics, hence, the decompression of the spring is 0.50 m.

Hope this Helps!!!

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weeeeeb [17]

Answer:

F= 17,075\ N

Explanation:

When the car is under an accelerating force and hits a tree, the instant force received by the tree is the same force that is accelerating the car.

The accelerating force can be calculated using Newton's second law:

F=m\cdot a

Where m is the mass of the car and a is the acceleration.

F=3,415\ kg\cdot 5\ m/s^2

\boxed{F= 17,075\ N}

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A 1. 30 kg block slides with a speed of 0. 855 m/s on a frictionless horizontal surface until it encounters a spring with a forc
boyakko [2]

a) For 0 compressions:

potential energy = U =  0 J

kinetic energy = k = 0.383 J

total mechanical energy = E = 0.383 J

b) For compression of 1 cm:

potential energy = U = 0.0228 J

kinetic energy = k = 0.155 J

total mechanical energy = E = 0.383 J

c) For compression of 2 cm :

potential energy = U = 0.1104 J

kinetic energy = k = 0.272 J

total mechanical energy = E = 0.383 J

d) For compression of 3cm:

potential energy  = U = 0.248 J

kinetic energy = k = 0.177 J

total mechanical energy = E = 0.383 J

<h3>Method for solving:</h3>

The equations for kinetic energy is:

k= 1/2*m*v^{2}

The equation for elastic potential energy is:

U= 1/2*ks*x^{2}

Where,

m= mass of the block

v= velocity

ks= spring constant

x= displacement of the spring

(a)when compression= 0 cm

U= 1/2*ks*v{2}

U= 1/2*552*(0)^{2}

 = 0 J

Kinetic energy:

k= 1/2*m*x^{2}

k= 1/2*(1.05)*(0.855)^{2}

k= 0.383 J

Mechanical energy:

E= k + U

E= 0.383+0

E= 0.383 J

There will be no work done by friction or any other dissipative force, hence this energy will be conserved, or it will remain constant (like air resistance). This indicates that only spring potential energy will be created from the kinetic energy (there is no thermal energy due to friction, for example).

(b) spring potential = ?

U= 1/2* 457 N/m*(0.01)^{2}

U= 0.0228 J

Since the mechanical energy must remain constant, we may calculate the kinetic energy using the mechanical energy equation:

E= k + U

0.383= k + 0.0228

k= 0.383 - 0.228

k= 0.155

(c)spring constant when x= 0.02

U= 1/2*552*(0.02)^{2}

U= 0.1104 J

Using the equation of mechanical energy:

E= k +U

0.383= k+ 0.1104

k= 0.383 - 0.1104

k= 0.272 J

(d) U= 1/2*552*(0.03)^{2}

U= 0.2484 J

E= 0.383 J

k = E - U

k= 0.383- 0.206

k= 0.177

To learn more about spring potential energy visit:

brainly.com/question/28168175

#SPJ4

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