A tourist of mass \(60.0 \mathrm{~kg}\) notices a chest with a short chain attached to it at the bottom of the ocean. Imagining the riches it could contain, he decides to dive for the chest. He inhales fully, thus setting his average body density to \(945 \mathrm{~kg} / \mathrm{m}^{3}\), jumps into the ocean (with saltwater density = \(1020 \mathrm{~kg} / \mathrm{m}^{3}\) ), grabs the chain, and tries to pull the chest to the surface. Unfortunately, the chest is too heavy and will not move. Assume that the man does not touch the bottom. a) Draw the man's free-body diagram, and determine the tension on the chain. b) What mass (in kg) has a weight that is equivalent to the tension force in part (a)? c) After realizing he cannot free the chest, the tourist releases the chain. What is his upward acceleration (assuming that he simply allows the buoyant force to lift him up to the surface)?

Step by step solution

01

Calculate the volume of the tourist

First, we will calculate the volume of the tourist. We can use the formula: $$ \text{Volume} = \frac{\text{Mass}}{\text{Density}} $$ $$ V = \frac{60.0 \thinspace kg}{945 \thinspace kg/m^3} = 0.06349 \thinspace m^3 $$

02

Calculate the buoyant force

Next, we need to calculate the buoyant force on the tourist. According to Archimedes' principle, the buoyant force is equal to the weight of the liquid displaced by the body. The formula for the buoyant force is: $$ F_b = \rho_{\text{fluid}} \times V \times g $$ Where \(\rho_{\text{fluid}}\) is the density of saltwater, \(V\) is the volume of the tourist, and \(g\) is the acceleration due to gravity (\(9.81 \thinspace m/s^2\)). $$ F_b = 1020 \thinspace kg/m^3 \times 0.06349 \thinspace m^3 \times 9.81 \thinspace m/s^2 = 637.7 \thinspace N $$

03

Calculate the weight of the tourist

Now, we will calculate the weight of the tourist using the formula: $$ W = m \times g $$ Where \(m\) is the mass of the tourist and \(g\) is the acceleration due to gravity. $$ W = 60.0 \thinspace kg \times 9.81 \thinspace m/s^2 = 588.6 \thinspace N $$

04

Determine the tension on the chain

Finally, we will determine the tension on the chain by finding the difference between the buoyant force and the weight of the tourist: $$ T = F_b - W $$ $$ T = 637.7 \thinspace N - 588.6 \thinspace N = 49.1 \thinspace N $$ The tension on the chain is 49.1 N. b) Determine the mass with equivalent weight:

05

Calculate the mass with equivalent weight

To find the mass with a weight equivalent to the tension force, we can use the formula: $$ m = \frac{W}{g} $$ Where \(W\) is the tension force (49.1 N) and \(g\) is the acceleration due to gravity (9.81 m/s^2). $$ m = \frac{49.1 \thinspace N}{9.81 \thinspace m/s^2} = 5.01 \thinspace kg $$ The mass with an equivalent weight to the tension force is 5.01 kg. c) Calculate the upward acceleration:

06

Calculate the net upward force

The net upward force on the tourist when the chain is released is equal to the buoyant force: $$ F_{\text{net}} = F_b = 637.7 \thinspace N $$

07

Calculate the upward acceleration

To calculate the upward acceleration, we will use Newton's second law of motion: $$ F = m \times a $$ Solving for acceleration, we have: $$ a = \frac{F}{m} $$ We will use the net upward force (637.7 N) and the mass of the tourist (60 kg) in this formula. $$ a = \frac{637.7 \thinspace N}{60 \thinspace kg} = 10.63 \thinspace m/s^2 $$ The upward acceleration of the tourist when he releases the chain is 10.63 m/s².

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