(a) The free-body diagram for the system (the block of ice) is shown below. Here

N

is the normal force exerted by the lake surface on the block of ice. Note that the applied force vector

F

is drawn with its tail at the system to easily visualize its components. Since we are interested in determining the magnitude of the normal force, we need to consider motion in the vertical direction when applying Newton's second law. Note that the acceleration in the vertical direction is zero as there is no motion in that direction.

Newton's second law for the vertical direction can therefore be written as follows.
(b) In this case, the pulling force is pointing up and to the right. The sine component of

F

is therefore directed upward now. As in part (a), the acceleration in the vertical direction is zero and Newton's second law can be written as follows.

(a) Since we have been asked to use energy content, we need to consider gravitational potential energy of the box, kinetic energy of the box, and elastic potential energy of the spring. Since gravitational and elastic potential energy are involved, we need to select a point for zero gravitational potential energy and a point for zero elastic potential energy. It would be advantageous if we could select the same point for zero gravitational potential energy and zero elastic potential energy. Let's agree to choose zero gravitational potential energy and zero elastic potential energy to be the point where the box makes initial contact with the spring.

As the box travels from position A to B, the only force doing work on the box is the force of gravity. Since this is a conservative force, energy is conserved and the total energy (potential plus kinetic) at point A is equal to the total energy at point B. This allows us to write Inserting details for the gravitational potential energy and kinetic energy, obtain where

d sin θ

is the initial vertical distance the box is above the level of zero gravitational potential energy expressed in terms of the distance up the incline and the angle of inclination. Solving this expression for the speed of the box at position B, obtain Entering values, we have Since we are interested in the speed, we will use the positive value.

(b) As the box travels from A to C, only conservative forces are doing work on the box. As a result, energy is conserved and the total energy at point A is equal to the total energy at point C. This allows us to equate the potential (gravitational) and kinetic energy at point A to the potential (gravitational and spring) and kinetic energy at point C. This may be written as Inserting details, obtain Solving for the spring constant, obtain Entering values, we have
(c) Only conservative forces act on the box; as a result, the total energy is a constant. The initial gravitational potential energy at A is traded for kinetic energy as the box travels from A to B, and then back into gravitational potential energy and spring potential energy as the box travels on to point C.

Note: We are displaying rounded intermediate values for practical purposes. However, the calculations are made using the unrounded values.

(a) The quantity of heat transferred is given by the equation

Q = nC_VΔT

where n is the number of moles of the gas, ΔT is the change in temperature of the gas, and

C_V

is the molar specific heat at constant volume.

Substitute the given values to get
(b) Let

E_int

represent the internal energy, Q the net heat transferred into the system and

W = −

	PΔV

the net work done on the system at constant pressure. Then the first law of thermodynamics can be written as Since the volume of the gas remains constant, the work done on the gas is zero. Therefore, we have
(c) The quantity of heat transferred is given by the equation

Q = nC_PΔT

where n is the number of moles of the gas, ΔT is the change in temperature of the gas, and

C_P

is the molar specific heat at constant pressure.

Substitute the given values to get Note that this quantity is greater than that in part (a). This is because some of the heat input goes into increasing the internal energy of the gas, and some goes into increasing the volume of the gas at a constant pressure (which means work is done by the gas on its surroundings as it heats up).

(d)

ΔE_int

is the same for the constant volume and constant pressure processes since the temperature change is the same.

Therefore

ΔE_int = 1.44 ✕ 10⁴ J.

We use

ΔE_int = Q + W

to find the work done on the gas. Alternatively, we can calculate work directly from the temperature change. Because the work on the gas is at a constant pressure, For an ideal gas, we can substitute

PV = nRT

into the above and obtain The work done on the gas is negative, because the volume of the gas is increasing. (The work done by the gas on the surroundings is the opposite sign of this result, and so would be positive in this case.)