Question

Suppose a 60.0-kg gymnast climbs a rope.

(a) What is the tension in the rope if he climbs at a constant speed?

(b) What is the tension in the rope if he accelerates upward at a rate of 1.50 m/s²?

Short answer

(a)The tension in the rope with constant speed is 588 N.

(b) The tension in the rope with some acceleration is 678 N.

Step-by-step solution

Theory

Apply Newton’s second law of motion.

\({F_{{\rm{net}}}} = ma\)

\(T - mg = ma\)……………….. (i)

Here, F_net is the net force, m is the mass of gymnast, T is the tension in the rope, g is the acceleration due to gravity, and a is the acceleration.

(a) Determine the tension in the rope when climbing at constant speed

At constant speed the acceleration is zero.

Substitute 60 kg for m, 9.8 m/s² for g, and 0 for a in equation (i) in the above expression, and we get,

\(\begin{array}{c}T - 60\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} = 0\\T = 588\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = 588\;{\rm{N}}\end{array}\)

Hence,the tension in the rope with constant speed is 588 N.

(b) Determine the tension in the rope with some acceleration

Substitute 60 kg for m, 9.8 m/s² for g, and 1.50 m/s² for a in equation (i) in the above expression, and we get,

\(\begin{array}{c}T - 60\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} = 60\;{\rm{kg}} \times 1.50\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = \left( {90 + 588} \right)\;{\rm{kg}} \cdot {\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = 678\;{\rm{N}}\end{array}\)

Hence, the tension in the rope with some acceleration is 678 N.