Is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity.

Thermal = 0.48 W

Explanation:

Given:

- The complete question is:

" Starting from rest, a 0.0367-kilogram steel ball sinks into a vat of corn syrup. The thick syrup exerts a viscous drag force that is proportional to the ball's velocity: where C = 0.270 N- s/m is a constant related to the size and composition of the ball as well as the viscosity of the syrup. Find the rate at which gravitational energy is converted to thermal energy once the ball reaches terminal velocity. After the ball reaches terminal velocity its kinetic energy is constant Any further work done by gravity is therefore directly transferred to the syrup and ball as thermal energy The rate at which the energy is converted is the power supplied by gravity: where F is the force, v is the speed of the ball and B is the relative angle between the force and velocity vectors. The terminal velocity of the ball can be determined from the given information using Newton's Laws. "

Solution:

- Once the terminal velocity Vt of the body is reached, it begins to move downward  at constant speed. The drag force Fd then attains a constant value:

                                         Fd = - C • Vt

- Instant the rate of conversion of gravitational potential energy to thermal  energy becomes equal to the power dissipated by the drag force at the terminal  velocity is given by:

                                        Fd • Vt = - C • Vt ²

- To get the general velocity V , solve the equation of motion:

                                        mv' = -m*g - C*V

                                        v' = -g - [C/ m] * V

Where,

                                        v' = dv / dt

                                        dv / dt =  -g - [C/ m] * V

- We will denote [C/ m]  as constant B:

                                        dv / dt =  -B*[ g/B +  V ]

- Separate variables:

                                        dv / [ g/B +  V ]  =  -B*dt

- Integrate both sides:

                                        Ln | g/B +  V | = -B*t + C

                                        g/B +  V = C*e^( -B*t )

                                        V = C*e^( -B*t ) - g/B

- The above expression for velocity gives the velocity of the steel ball at any time t. We know that terminal velocity Vt is achieved when t - > ∞. Then we have:

                                       Vt = 0 - g/B = -m*g/C

- Then from power equation we have:

                                        P = - C*Vt^2

                                        P = -C ( -m*g/C )^2

                                        P = -(mg)^2 / C

                                        P = -(0.0367*9.81)^2 / 0.27

                                        P = - 0.48 W

- The rate at which gravitational energy is lost to surrounding gives rise to thermal energy of surrounding fluid Thermal = 0.48 W .

what was the acceleration of the 2 kg cart?

⇒ 16.0 cm/s2

what was the acceleration of the 4 kg cart?

⇒ 8.1 cm/s2

what was the acceleration of the 6 kg cart?

⇒ 5.3 cm/s2