An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g 10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

An object with mass M is attached to the end of a string and is raised vertically at a constant acceleration of g/10 . If it has been raised a distance ℓ from rest, how much work has been done by the tension in the string?

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Given:

Mass of the object = M

Acceleration of the object = g/10

Distance = ℓ

Asked:

Work by Tension = W = ?

Solution:

Let's find the magnitude of tension as follows:

[tex]\Sigma F = ma[/tex]

[tex]T - Mg = Ma[/tex]

[tex]T = Mg + Ma[/tex]

[tex]T = M(g + a)[/tex]

[tex]T = M(g + \frac{1}{10}g)[/tex]

[tex]T = M(\frac{11}{10}g)[/tex]

[tex]T = \frac{11}{10}Mg[/tex]

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[tex]W = T \times L[/tex]

[tex]W = \frac{11}{10}Mg \times L[/tex]

[tex]W = \frac{11}{10}MgL[/tex]

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The work by tension is ¹¹/₁₀ Mgℓ

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Answer details

Grade: High School

Subject: Physics

Chapter: Dynamics

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

snehashish65 snehashish65 · Answer 2 · 2021-10-11T12:36:22+02:00

Answer:

The work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

Explanation:

Given data:

Mass of object is, M.

Acceleration of object is, a = g/10.

Distance covered vertically is, L.

The work done by tension in the string is given as,

[tex]W = T \times L[/tex] .......................................................... (1)

Here, T is the tension force on string.

Apply the equilibrium of forces on string as,

[tex]T- Mg=Ma[/tex]

Here, g is gravitational acceleration.

[tex]T- Mg=M(\dfrac{g}{10} )\\T=M(\dfrac{g}{10} )+Mg\\T=\dfrac{11}{10}Mg[/tex]

Substituting value in equation (1) as,

[tex]W = \dfrac{11}{10}Mg \times L\\W = \dfrac{11}{10}MgL}[/tex]

Thus, the work done by tension in string is [tex]\dfrac{11}{10}MgL}[/tex].

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