Question

31. (a) The masses M in the two figures have identical values. Which of the twosystems has a smaller gravitational potential energy? Why?MM-fig: (b)fig: (a)​

Answer 1

Explanation:

Work Done Against Gravity

Climbing stairs and lifting objects is work in both the scientific and everyday sense—it is work done against the gravitational force. When there is work, there is a transformation of energy. The work done against the gravitational force goes into an important form of stored energy that we will explore in this section.

(a) The weight attached to the cuckoo clock is raised by a height h shown by a displacement vector d pointing upward. The weight is attached to a winding chain labeled with a force F vector pointing downward. Vector d is also shown in the same direction as force F. E in is equal to W and W is equal to m g h. (b) The weight attached to the cuckoo clock moves downward. E out is equal to m g h.

Figure 1. (a) The work done to lift the weight is stored in the mass-Earth system as gravitational potential energy. (b) As the weight moves downward, this gravitational potential energy is transferred to the cuckoo clock.

Let us calculate the work done in lifting an object of mass m through a height h, such as in Figure 1. If the object is lifted straight up at constant speed, then the force needed to lift it is equal to its weight mg. The work done on the mass is then W = Fd = mgh. We define this to be the gravitational potential energy (PEg) put into (or gained by) the object-Earth system. This energy is associated with the state of separation between two objects that attract each other by the gravitational force. For convenience, we refer to this as the PEg gained by the object, recognizing that this is energy stored in the gravitational field of Earth. Why do we use the word “system”? Potential energy is a property of a system rather than of a single object—due to its physical position. An object’s gravitational potential is due to its position relative to the surroundings within the Earth-object system. The force applied to the object is an external force, from outside the system. When it does positive work it increases the gravitational potential energy of the system. Because gravitational potential energy depends on relative position, we need a reference level at which to set the potential energy equal to 0. We usually choose this point to be Earth’s surface, but this point is arbitrary; what is important is the difference in gravitational potential energy, because this difference is what relates to the work done. The difference in gravitational potential energy of an object (in the Earth-object system) between two rungs of a ladder will be the same for the first two rungs as for the last two rungs.

Converting Between Potential Energy and Kinetic Energy

Gravitational potential energy may be converted to other forms of energy, such as kinetic energy. If we release the mass, gravitational force will do an amount of work equal to mgh on it, thereby increasing its kinetic energy by that same amount (by the work-energy theorem). We will find it more useful to consider just the conversion of PEg to KE without explicitly considering the intermediate step of work. (See Example 2.) This shortcut makes it is easier to solve problems using energy (if possible) rather than explicitly using forces.

More precisely, we define the change in gravitational potential energy ΔPEg to be ΔPEg = mgh, where, for simplicity, we denote the change in height by h rather than the usual Δh. Note that h is positive when the final height is greater than the initial height, and vice versa. For example, if a 0.500-kg mass hung from a cuckoo clock is raised 1.00 m

Answer 2

The second system has smaller gravitational potential energy.

Explanation:

The missing figure is attached.

We know that

Gravitational potential energy between two masses M and m, separated by a distance r is given by

$U=-\frac{GMm}{r}$

As shown, in figure (a), the distance between two masses M is

$\sqrt{r^2+r^2}=\sqrt{2r^2}=r\sqrt{2}$

Therefore the gravitational  potential energy is

$U=-\frac{GM^2}{r\sqrt{2}}$

In figure (b), the distance between the masses is

$2r$

Therefore the gravitational  potential energy is

$U'=-\frac{GM^2}{2r}$

Since $2r>r\sqrt{2}$

And the gravitational  potential energy has a negative sign

Therefore, in second case, the gravitational  potential energy will be smaller.

Hope this answer is helpful.

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