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A gymnast of mass 69.0 kg hangs from a vertical rope attached to the ceiling. You can ignore the weight of the rope and assume that the rope does not stretch. Use the value 9.81 m/s2 for the acceleration of gravity.Calculate in Newtons:A) The tension T in the rope if the gymnast hangs motionless on the rope.B) The tension T in the rope if the gymnast climbs the rope at a constant rate.C) The tension T in the rope if the gymnast climbs up the rope with an upward acceleration of magnitude 0.700 m/s2.D) The tension T in the rope if the gymnast slides down the rope with a downward acceleration of magnitude 0.700 m/s2.

Respuesta :

Answer:

Part a)

[tex]T = 676.9 N[/tex]

Part b)

[tex]T = 676.9 N[/tex]

Part c)

[tex]T = 725.2 N[/tex]

Part d)

[tex]T = 628.6 N[/tex]

Explanation:

Part a)

Since the gymnast is holding the rope and remains stationary

So total force is balanced on it

so we will have

T = mg

[tex]T = (69)(9.81)[/tex]

[tex]T = 676.9 N[/tex]

Part b)

Now gymnast start climbing the rope at uniform speed

so here we have

[tex]a = 0[/tex]

so again we will have same force equation

[tex]T = mg[/tex]

[tex]T = 676.9 N[/tex]

Part c)

Now gymnast start climbing up the rope at uniform acceleration

so now we will have

[tex]T - mg = ma[/tex]

[tex]T = mg + ma[/tex]

[tex]T = 69(9.81 + 0.700)[/tex]

[tex]T = 725.2 N[/tex]

Part d)

Now gymnast start sliding downwards with same uniform acceleration

so now we will have

[tex]mg - T = ma[/tex]

now we have

[tex]mg - ma = T[/tex]

[tex]T = 69(9.81 - 0.7)[/tex]

[tex]T = 628.6 N[/tex]

A) The tension T in the rope if the gymnast hangs motionless on the rope is 676.9N

B) The tension T in the rope if the gymnast climbs the rope at a constant rate is 676.9N

C) The tension T in the rope if the gymnast climbs up the rope with an upward acceleration is 725.2N

D) The tension T in the rope, if the gymnast slides down the rope with a downward acceleration, is 628.6N

Given that the mass of the gymnast is m = 69kg

Resolving the forces:

(a) The gymnast is holding the rope and is in equilibrium

so

[tex]T=mg\\\\T=69*9.8=676.9N[/tex]

where T is the tension in the rope and mg is the weight

(b) The gymnast starts climbing the rope at uniform speed so there is no acceleration.

Therefore

[tex]T=mg+m*0\\\\T=69*9.8=676.9N[/tex]

(c) The gymnast starts climbing up the rope at uniform acceleration so there will be a pseudo force acting downwards along the weight,

hence, the tension in the rope:

[tex]T=mg+ma[/tex]

where a = 0.7m/s²

[tex]T==69*9.8+60*0.7=725.2N[/tex]

(d) The gymnast starts sliding downwards with the same uniform acceleration

so there will be a pseudo force acting upwards opposite to the weight

therefore,

[tex]T=mg-ma=69*9.8-69*0.7=628.6N[/tex]

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