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Bond [772]
3 years ago
6

If it requires 2.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re

quired to stretch it an additional 4.0 cm?
Physics
1 answer:
valina [46]3 years ago
7 0

Answer:

16 J

Explanation:

It is given that,

Work done, W = 2 J

A spring is stretched by 2.0 cm from its equilibrium length

We need to find how much more work will be required to stretch it an additional 4.0 cm.

Let k is the spring constant of the spring. When W = 2J, and x = 2 cm, then energy required to stretch the spring is :

U=\dfrac{1}{2}kx^2\\\\k=\dfrac{2U}{x^2}\\\\k=\dfrac{2(2)}{(0.02)^2}\\\\k=10000\ N/m

The energy required to stretch the spring from 2 cm to additional 4 cm i.e. 2+4= 6 cm.

W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\=\dfrac{1}{2}\times 10000\times ((0.06)^2-(0.02)^2)\\\\W=16\ J

So, the required work done is 16 J.

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Answer:

Lens

Explanation:

Lenses fit this description.

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3 years ago
in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo
guajiro [1.7K]

Explanation:

The momentum of the three objects are as follow :

11 kg-m/s, -65 kg-m/s and -100 kg-m/s

Before collision, the momentum of the system is :

P_i=11+(-65)+(-100)\\\\P_i=-154\ kg-m/s

After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.

It would mean that, after collision, momentum of the system is equal to the initial momentum.

Hence, final momentum = -154 kg-m/s.

4 0
3 years ago
A block with mass m = 0.450 kg is attached to one end of an ideal spring and moves on a horizontal frictionless surface. The oth
svetoff [14.1K]

Answer:

k = 26.25 N/m

Explanation:

given,

mass of the block= 0.450

distance of the block = + 0.240

acceleration = a_x = -14.0 m/s²

velocity = v_x = + 4 m/s

spring force constant (k) = ?

we know,

x = A cos (ωt - ∅).....(1)

v = - ω A cos (ωt - ∅)....(2)

a = ω²A cos (ωt - ∅).........(3)

\omega = \sqrt{\dfrac{k}{m}}

now from equation (3)

a_x = \dfrac{k}{m}x

k = \dfrac{m a_x}{x}

k = \dfrac{0.45 \times (-14)}{0.24}

k = 26.25 N/m

hence, spring force constant is equal to k = 26.25 N/m

8 0
3 years ago
Read 2 more answers
Three strings, attached to the sides of a rectangular frame, are tied together by a knot as shown in the figure. The magnitude o
Anna [14]

The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

Learn more about Tension here brainly.com/question/2287912

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8 0
2 years ago
Pweese help luvs qwq<br> pweese look at the image?
hjlf

Answer:

12 m/s

Explanation:

divide distance over time

72/6 = 12

6 0
2 years ago
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