Suppose a 60.0-kg gymnast climbs a rope. (a) what is the tension in the rope if he climbs at a constant speed? (

Constant speed means no acceleration. In terms of the tension this means there's no additional forces on the rope to account for the accelerating person. So the only force felt by the rope is the weight which is the same whether they are climbing or just hanging from the rope.
So tension is
T=mg=(60)(9.81)=588.6N

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Nudrat Nudrat · Answer 2 · 2019-09-04T12:49:07+02:00

The tension in the rope when gymnast climbs a rope with constant speed will be \boxed{588{\text{ N}}}.

Further explanation:

We have to calculate the tension in the rope when the gymnast climbs a rope with the constant speed.

Given:

We have given that,

The mass of the gymnast is [tex]60{\text{ kg}}[/tex].

Formula and concept used:

Consider a body is moving with an acceleration in upward direction, then the direction of the inertia force will be in the opposite direction of the direction of the acceleration.

That is, the direction of the inertia force will be in the downward direction for the moving body.

Inertia force can be calculated as,

[tex]\boxed{{F_I}=ma}[/tex]

Here, [tex]{F_I}[/tex] is the inertia force and is the acceleration.

So, the net downward force will be equal to the sum of the weight of the body and the inertia force.

[tex]\boxed{F=W + {F_I}}[/tex]

Substitute the values of [tex]W[/tex] as [tex]mg[/tex] and [tex]{F_I}[/tex] as [tex]ma[/tex] in above equation.

[tex]\boxed{F=mg + ma}[/tex] …… (1)

Here, gymnast is climbing with the constant speed.

So, acceleration of the gymnast in upward direction will be zero.

Substitute [tex]0[/tex] for [tex]a[/tex] in equation (1).

So, the net downward force will be,

[tex]F = mg[/tex]

This downward force will be balanced by the tension in the rope.

So, tension in the rope will be,

[tex]\boxed{T = mg}[/tex] …… (2)

Calculation:

Substitute the value [tex]60{\text{ kg}}[/tex] for [tex]m[/tex] and [tex]9.8{\text{ }}{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^2}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^2}}}[/tex] for [tex]g[/tex] in equation (2).

[tex]\begin{aligned}T&=60\times9.8\\&=588{\text{N}}\\\end{aligned}[/tex]

So, the tension in the rope will be [tex]\boxed{588{\text{ N}}}[/tex].

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Answer details:

Grade: High School

Subject: Physics

Chapter: Force

Keywords:

Astronaut, climb a rope, tension in the rope, constant speed, acceleration, acceleration due to gravity, force, weight, mass, downward, upward, direction, motion.