Question

A horse whose mass is M gallops at constant speed v up a long hill whose vertical height is h, taking an amount of time t to reach the top. The horse's hooves do not slip on the rocky ground, so the work done by the force of the ground on the hooves is zero (no displacement of the force). When the horse started running, its temperature rose quickly to a point at which from then on, heat transferred from the horse to the air keeps the horse's temperature constant.(a) First consider the horse as the system of interest. In the initial state the horse is already moving at speed v. In the final state the horse is at the top of the hill, still moving at speed v. Write out the energy principle?Esys = W + Qfor the system of the horse alone. The terms on the left hand side should include only energy changes for the system, while the terms on the right hand side should relate to the surroundings (everything else).Which of the following terms are equal to 0?1W?Echemical,horse?(mhorsec2)?Khorse?Ethermal,horseQ

Answer 1

Answer:

The answer to the question is;

the following terms on the left hand side of the energy equation which are  changes in the energy for the system are equal to 0;

• Δ$E_{Thermal}$ = 0

• Δ$m_{horse}c^{2}$   = 0

• ΔK$_{horse}$ = 0.

Explanation:

To solve the question we note that the final form is

E$_{sys}$ = W +Q

Since we are required to write out the system for only the hose, we have

Δ$E_{System}$ = Δ$m_{horse}c^{2}$ + ΔK$_{horse}$ + Δ$E_{Chemical}$ + Δ$E_{Thermal}$  = W + Q = -M·g·h - Q Abs

Since we have that the horses temperature is constant, we have

Δ$E_{Thermal}$  = 0

Again the speed v of the horse remain constant ∴ Δ$m_{horse}c^{2}$  =0

Also ΔK$_{horse}$  = 0  as the horse is moving at constant speed =0

That is the following terms are equal to 0

Δ$E_{Thermal}$, Δ$m_{horse}c^{2}$ , ΔK$_{horse}$.

Answer 2

A horse whose mass is M gallops at constant speed v up a long hill whose vertical height is h, taking an amount of time t to reach the top. The horse's hooves do not slip on the rocky ground, so the work done by the force of the ground on the hooves is zero. Thus, the energy principle for the system can is:  $\mathbf{\Delta E_{horse} = \Delta E_{chemical} + \Delta E_{thermal}+ \Delta K}$ and the following terms that are equal to 0 are  $\mathbf{\Delta E_{thermal \ horse } =0 }$ and $\mathbf{\Delta K_{horse}=0}$

           From the given information, we are to consider the horse to be the system. Since the horse is regarded as the system, then the energies flowing in the system are:

• chemical energy (provided that the horse happens to be a living organism)
• thermal energy (as a result of variations in the horse's temperature)
• kinetic energy (due to variation of the horse's speed)

           In the system, we cannot assert that the horse possesses gravitational potential energy. This is because, for a system to have gravitational potential energy, there needs to be the existence of two objects for the gravitational attraction to coexist within the system.

Hence, the energy principle for the system can be computed as:

• $\mathbf{\Delta H_{horse} = \Delta E_{chemical}+ \Delta E_{thermal}+\Delta \ K}$

           Recall from the question that, the horse possesses the ability to keep and maintain its temperature constant through heat transfer with the surrounding air.

• Thus, the $\mathbf{\Delta E_{thermal }=0}$

           However, since the speed is also said to be constant;

• $\mathbf{\Delta K = 0 }$

           Now, from the given options.

• $\mathbf{\Delta E_{thermal.horse} = 0 \ and \ \Delta K_{horse} = 0}}$

Therefore, we can conclude that the total change in the horse energy is brought about by the changes in its chemical energy.

Learn more about the energy principle here:

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