Question

Question:(II) A 62-kg trampoline artist jumps upward from the top of a platform with a vertical speed of\({\bf{5}}{\bf{.0 m/s}}\)(a) How fast is he going as he lands on the trampoline, 2.0 m below (Fig. 6–40)? (b) If the trampoline behaves like a spring of spring constant\({\bf{5}}{\bf{.8 \times 1}}{{\bf{0}}^{\bf{4}}}{\bf{ N/m}}\)how far down does he depress it?

FIGURE 6–40Problem 38.

Short answer

(a) The artist is going to land on the trampoline at a speed of 7.71 m/s.

(b) The compression distance of the spring is 0.252 m.

Step-by-step solution

Understanding the motion of the trampoline artist

The trampoline artist initially jumps upward from the platform. The final velocity of the artist can be evaluated with the help of Newton’s third law.

When the artist jumps on the trampoline, it behaves like a spring. From the energy conservation theory, the artist's kinetic energy is equal to the spring potential energy. The spring displacement can then be determined.

Identification of given data

The given data can be listed below as:

• The mass of the trampoline artist is\(m = 62{\rm{ kg}}\).
• The vertical speed of the artist is\(u = 4.5{\rm{ m/s}}\).
• The distance between the trampoline and platform is\(y = 2{\rm{ m}}\).
• The trampoline behaves like a spring, so the spring’s stiffness is\(k = 5.8 \times {10^4}{\bf{ }}{\rm{N/m}}\).
• The acceleration due to gravity is\(g = 9.81{\rm{ m/}}{{\rm{s}}^2}\).

Representation of the diagram of the trampoline

The diagram can be shown as:

Here, \({y_1}\) is the spring displacement when the trampoline behaves like a spring.

(a) Determination of the speed of the artist just before landing on the trampoline

From Newton’s equation of motion, the final velocity of the artist just before he hits the trampoline can be expressed as:

\({v^2} = {u^2} + 2ay\)

Substitute the values in the above expression.

\(\begin{array}{c}{v^2} = {\left( {4.5{\rm{ m/s}}} \right)^2} + 2 \times 9.81{\rm{ m/}}{{\rm{s}}^2} \times 2{\rm{ m}}\\v = \sqrt {59.49} {\rm{ m/s}}\\ = 7.71{\rm{ m/s}}\end{array}\)

Thus, the artist is going to land on the trampoline at a speed of 7.71 m/s.

(b) Determination of the trampoline’s compressed distance

The initial point is considered when the artist is just about to hit the trampoline. The final point is considered when the trampoline is compressed and the artist comes to the rest position.

The total initial energy is equal to the total final energy. The energy can be expressed as:

\(\begin{array}{c}P.{E_i} + K.{E_i} = P.{E_f} + K.{E_f} + S.{E_f}\\0 + \frac{1}{2}m{v^2} = 0 + \frac{1}{2}ky_1^2 + 0\\m{v^2} = ky_1^2\\y_1^2 = \frac{{m{v^2}}}{k}\end{array}\)

Here,\(v\)is the velocity of the artist. The initial potential energy\(P.{E_i}\)is equal to zero. The gravitational potential energy is zero because the distance from the reference position is almost the same as the final position.

\(K.{E_i}\)is the initial kinetic energy of the artist,\(P.{E_f}\)is the final gravitational potential energy, which is considered zero,\(S.{E_f}\)is the potential energy of the spring when the trampoline is compressed, and\(K.{E_f}\)is the final kinetic energy after the artist jumps on the trampoline, which is equal to zero.

Substitute the values in the above expression.

\(\begin{array}{c}{\left( {{y_1}} \right)^2} = \frac{{{\rm{62 kg}} \times {{\left( {7.71{\rm{ m/s}}} \right)}^2}}}{{\left( {5.8 \times {{10}^4}{\bf{ }}{\rm{N/m}}} \right)\left( {\frac{{1{\rm{ kg/}}{{\rm{s}}^2}}}{{1{\rm{ N/m}}}}} \right)}}\\y = 0.252{\rm{ m}}\end{array}\)

Thus, the compressed distance of the trampoline is 0.252 m.