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It requires a force of 100 N to stretch a spring of negligible mass by a distance of 0.1 m. If the spring instead stretches a distance of 0.25 m, how much force was applied to it?

answer choices
1. 200 N
2. 300 N
3. 250 N
4. 100 N
5. 175 N
6. 325 N
7. 150 N
8. 225 N
9. 275 N

Respuesta :

by the formula of spring force we know that

[tex]F = kx[/tex]

here we know that

[tex]F = 100 N[/tex]

[tex]x = 0.1 m[/tex]

now we will have

[tex]100 = k \times 0.1 [/tex]

[tex]k = 1000 N/m[/tex]

now by similar way if the stretch in spring is 0.25 m

force is given by

[tex]F = kx[/tex]

[tex]F = 1000 \times 0.25[/tex]

[tex]F = 250 N[/tex]

so it will require F = 250 N force

Answer: The force applied on the spring is 250 N

Explanation:

To calculate the spring constant, we use the equation:

[tex]F=kx[/tex]       ......(1)

where,

F = force exerted on the spring = 100 N

k = spring constant = ?

x = length of the spring = 0.1 m

Putting values in equation 1, we get:

[tex]100N=k\times 0.1m\\\\k=\frac{100N}{0.1m}=1000N/m[/tex]

Now, calculating the force exerted on the spring, we use equation 1:

We are given:

[tex]k=1000N/m\\x=0.25m[/tex]

Putting values in above equation, we get:

[tex]F=1000N/m\times 0.25m\\\\F=250N[/tex]

Hence, the force applied on the spring is 250 N

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