The springs used in ball-point pens are some of the weaker examples found in practical applications. It usually only takes about 0.005 kilograms of mass resting on such a spring to compress it by 0.001 meters. Which of the following most closely describes the force constant of a typical ball-point pen spring?
A 0.05 N/m
B. 0.5 N/m
C. 5.0 N/m
D. 50 N/m
E. 500 N/m

Answers

Answer 1

The force-constant of a typical ball-point pen spring is approximately 50 N/m. Correct option is D. 50 N/m

The force constant of a spring can be calculated using Hooke's-Law, which states that the force exerted by a spring is directly proportional to its displacement.

F ∝ x ⇒ F = -kx

Here, F is the force, k is the force constant, and x is the displacement.

Here in this case, the displacement given is x = 0.001 m , the mass resting is 0.005 kg. i.e. m = 0.005 kg.

The force will be F = mg = 0.005 × 9.8 = 0.049 N

If we solve this,

0.049 N = -k × 0.001 m

k = 0.049 N / 0.001 m = 49 N/m

The approximation we get is 49 N/m and the option near this value id option D. Therefore, the force constant of a typical ball-point pen spring is approximately 50 N/m.

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Related Questions

which one of the following sets of quantum numbers can correctly represent a 3p orbital? group of answer choices n

Answers

Answer:
d) (n = 3, 1 = 1, mi = -1)
Explanation:
A 3p orbital must have n = 3.
That eliminates Option (a), because it has n = 1.
A p orbital must have I = 1.
That eliminates Options (b), (c), and (e), because
they have I = 3, 2, and O.
The only correct option is
n = 3, 1=1, mi =-1

Example 2.2 (Lecture Notes)
Three coplanar forces of
magnitudes 80 N, 50 N and
60 N act a point O as shown
in the figure. Determine the
resultant of these force.
A
C
50 N
达
20
60⁰
80 N
60 N
Basic Mechanics Lecture Notes Prepared By E. Adaze, Lecturer, Dept. of Mech. Eng., UMAT, Tarkwa 2021
B
16
Resolution of Forces into Components

Answers

There is no consequence when a point experiences three or more coplanar forces and the vector diagram closes. The forces affecting the point are balanced.

How do you calculate the coplanar forces' resultant force?

On the other hand, the Pythagorean Theorem, F = SQRT (Fx2 + Fy2), can be used to get the consequent if the magnitude of the rectangular components is known. Knowing that the tan is equal to the opposite side of the adjacent side, one can use trigonometry to calculate the direction of the resultant.

Let us resolve each of the forces into components along the x and y axes.

For the 80 N force:

Fx = 80 N cos(60°) = 80 N x 0.5 = 40 N

Fy = 80 N sin(60°) = 80 N x 0.866 = 69.28 N

For the 50 N force:

Fx = 50 N cos(20°) = 50 N x 0.94 = 47 N

Fy = 50 N sin(20°) = 50 N x 0.342 = 17.10 N

For the 60 N force:

Fx = 60 N cos(60°) = 60 N x 0.5 = 30 N

Fy = 60 N sin(60°) = 60 N x 0.866 = 51.96 N

Resultant Force:

The resultant force is then determined by adding the components:

Fx = 40 N + 47 N + 30 N = 117 N

Fy = 69.28 N + 17.10 N + 51.96 N = 138.34 N

Therefore, the magnitude of the resultant force is:

|F| = √(Fx2 + Fy2) = √(117 x 117 + 138.34 x 138.34) = 184.2 N

The direction of the resultant force is:

θ = tan−1(Fy/Fx) = tan−1(138.34/117) = 55.7°

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