Knowing that the kinetic energy of an object may be determined by

K = (1/2)mv²

and that the two balls have the same kinetic energy, we may write

K_bb = K_pp or (1/2)m_bbv_bb² = (1/2)m_ppv_pp².

Solving for the speed of the ping-pong ball, obtain: Just for comparison, the speed of sound in air at room temperature is about 340 m/s.

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

(a) We have been asked to use energy considerations to determine the speed of the objects after they have traveled a distance h. Since the masses move in the gravitational field of the Earth, we should expect to use both the work done by the force of gravity and kinetic energy. Before moving on, let's recognize that m₁ moves downward at the same speed m₂ moves upward. Since m₁ and m₂ have the same speed, there is no need for subscripts and we may write

v₁ = v₂ ≡ v.

Since the tension is an internal force, it does no net work on the two masses. The only force contributing to the net work done on the two masses is the force of gravity. We may write the net work done on the two masses by the force of gravity as where

W_g1

and

W_g2

represent the work done by gravity on masses m₁ and m₂, respectively;

F_g1 = m₁g

and

F_g2 = m₂g

represent the force of gravity acting on m₁ and m₂, respectively; h represents the distance the two masses travel; 0° is the angle between the direction of the force of gravity acting on m₁ and the displacement vector for m₁; 180° is the angle between the direction of the force of gravity acting on m₂ and the displacement vector for m₂; g is the magnitude of the acceleration due to gravity. Noting that the initial kinetic energy is zero and using the work-energy theorem, we may write the final kinetic energy as Solving for v we obtain Inserting the relationship between the masses, obtain Simplifying by canceling the mass, obtain Finally, inserting values we have Since we are interested in the speed, we have chosen the positive value.

(b) Next, we need to find the magnitude of the tension in the cord. Since the net work done on the system

(m₁, m₂,

and the pulley) by the tension in the cord is zero, in order to include the work done by the tension we will need to look at one mass or the other, let's choose m₁. The two forces acting on m₁ are the tension in the cord and the force of gravity. Since the tension is a nonconservative force, any work done by the tension on m₁ is responsible for the change in energy of m₁. This may be written as The work done by the tension on m₁ may also be expressed as

W_T1 = Th cos 180° = −Th,

where T represents the tension in the cord, h represents the distance the mass travels while acted on by the tension, and 180° is the angle between the direction of the tension acting on m₁ and the displacement of m₁. Equating these two expressions for the work done by the tension on m₁, obtain Solving for the tension we have Inserting values, we obtain:

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

(a) The net force acting on the crate may be determined by The negative sign informs us the net force on the crate is in a direction opposite the motion of the crate. The magnitude of the net force acting on the crate is 27.2 N and this force opposes the motion of the crate.

(b) The net work may be determined by
(c) The work-energy theorem tells us that the net work done on the crate is equal to the change in kinetic energy of the crate. We can use this to determine the final speed of the crate. Inserting values obtain:

(a) For a conservative force, the work done by the force acting over a distance is the negative of the change in potential energy associated with that force. This allows us to write

W_g = −ΔPE_g = −(mgΔy) = −mg(−h) = mgh.

(b) According to the work−energy theorem, the net work done on an object is equal to the change in its kinetic energy. This allows us to write

ΔK = W_net = W_g = mgh.

(c) Since energy is conserved,

ΔK = −ΔPE,

or

K_f − K_i = −(−mgh)

or

(d) Notice that the answer to parts (a) and (b) depend on the change in gravitational potential energy. The answer to part (c) depends on the initial kinetic energy and also the change in gravitational potential energy. Since neither the launch speed or the change in potential energy depend on the launch angle, none of the answers are changed if the initial angle is changed.

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

(a) The only external force doing work on the system (m₁ and m₂) is the force of gravity and since this is a conservative force, energy is conserved. Knowing energy is conserved, we may write

ΔE = ΔKE + ΔPE = 0,

or

ΔKE = −ΔPE.

The change in kinetic energy may be expressed as Inserting expressions for the kinetic energy of each object, obtain: The change in gravitational potential energy may be expressed as Inserting the expressions for the change in kinetic and potential energy into the expression

ΔKE = −ΔPE,

obtain

(m₁ + m₂)v²/2 = −(m₁ − m₂)gh.

Solving this for v obtain
(b) Values are inserted in order to obtain a numerical answer.