Using Wext = ΔEsys when Wext does not Equal Zero

Guide 8-5. Using W_ext = ΔE_sys when W_ext does not Equal Zero

Up to this point, you've been solving problems for which all the forces were conservative. By taking your system to include all objects interacting through conservative forces, you were able to say mechanical energy was conserved. In that case, the general conservation of energy equation W_ext= ΔE_sys reduced to 0 = ΔE_sys. In the event that there are forces external to the system that do work, the left-hand side will no longer be 0. Any force that is external to the system and does work on the system will be represented in W_ext.

Here's a situation with an external force that does work on the system. Open this animation. A block is initially given a push to start it moving upward between two rails. The rails exert kinetic friction force on the block. We take the system to include the block and the Earth. Kinetic friction is an external force that does work on the system. Note that the U_g = 0 level is set at the starting position. Also note the following.

At t = 0 (Figure 1), all of the system energy is kinetic.
Figure 2 shows that as the block rises and slows down, U_g increases while K decreases. The system energy, however, isn't conserved. That's because the work done by the friction force on the block is negative, since the friction force is always opposite the displacement. Note also that the change in E_sys, which is negative, is equal to W_f. Therefore, W_ext = W_f= ΔE_sys as expected.
At the highest position (Figure 3), the kinetic energy is 0, since the block stops momentarily. In this situation, all of the system energy is gravitational potential. However, the system energy has continued to decrease while the work done by friction has become more negative.
When the block returns to its starting position (Figure 4), the gravitational potential energy has returned to 0. This is because the net displacement in the round trip is 0. The work done by friction, however, continued to become more negative even after the block reversed directions. This is because friction is not a conservative force.

Figure 1	Figure 2	Figure 3	Figure 4

t = 0	partway up	highest position	back to the starting position

Now let's look at some worked-out examples.

Example 1. A 17-g leaf falls from a tree to the ground. If the leaf was 5.30 m above the ground as it left the tree and its speed just before landing was 1.3 m/s, how work did air friction do on the leaf?

Given: v_i = 0 v_f = 1.3 m/s y_i = 5.3 m y_f = 0 m = 0.017 kg Goal: Find the work done by air friction on the leaf. System: leaf, Earth External forces: air friction initial state: leaf detaches from tree final state: leaf reaches ground Energy changes: ΔU_g is negative ΔK is positive	The direction of +y is defined to be up. The lowest position of the leaf is designated as y_f= 0.

	Write the general conservation of energy law.
	W_ext is W_air, because the force of air friction does work on the leaf. The energy changes in the system are kinetic and gravitational.
	Expand the delta notation in terms of initial and final quantities.
	The initial kinetic energy and final gravitational energy are 0.
	Substitute expressions for the remaining energy terms.
	Substitute the givens with units.
	The work done by air friction is negative. This is because the force of air friction on the leaf is opposite the direction of the leaf's displacement.

Example 2. The problem is 8-34, part a, on page 235 of the text. A 42.0-kg seal at an amusement park slides from rest down a ramp into the pool below. The top of the ramp is 1.75 m higher than the surface of the water and the ramp is inclined at an angle of 35.0° above the horizontal. If the seal reaches the water with a speed of 4.40 m/s, what is the work done by kinetic friction?

Given: v_i = 0 v_f = 4.40 m/s y_i = 1.75 m y_f = 0 θ = 35° m = 42.0 kg Goal: Find the work done by kinetic friction. System: seal, Earth External forces: normal, kinetic friction initial state: highest position of seal final state: lowest position of seal Energy changes: ΔU_g is negative since the seal decreases in elevation; ΔK is positive since the seal speeds up	The direction of +y is defined to be up. The lowest position of the seal is designated as y_f= 0.
	.
	Write the general conservation of energy law.
	W_ext = W_N + W_f, because the normal force and kinetic friction are both external forces. However W_N = 0, because the normal force is perpendicular to the displacement.
	The energy changes in the system are kinetic and gravitational.
	Expand the delta notation in terms of initial and final quantities.
	The initial kinetic energy and final gravitational energy are 0.
	Substitute expressions for the remaining energy terms.
	Substitute the givens with units.
	Note the work done by friction is negative as expected.

Example 3. A 64-kg jogger is running at a speed of 1.1 m/s as he reaches the bottom of a hill. He runs up the hill and has a speed of 3.5 m/s at the top of the hill. Air friction did a total amount of work of -3500 J on the jogger while he was running up the hill. The hill is 21 m high. How much work did the ground do on the jogger as he was climbing the hill?

Given: v_i = 1.1 m/s v_f = 3.5 m/s y_i = 0 m y_f = 21 m m = 64 kg W_f = -3500 J Goal: Find the work done by the ground on the jogger. System: hill, Earth External forces: normal, air friction, ground pushing jogger forward initial state: lowest position of jogger final state: highest position of jogger Energy changes: ΔU_g is positive since the jogger increases in elevation; ΔK is positive since the jogger speeds up	The direction of +y is defined to be up. The lowest position of the jogger is designated as y_i= 0.
	.
	Write the general conservation of energy law.
	W_ext is composed of three terms. The normal force does 0 work on the jogger. Air friction does negative work on the jogger. By Newton's 3rd law, the force of the jogger's feet backward on the ground is equal and opposite to the force of the ground forward on the jogger's feet. The latter force does positive work on the jogger.
	The energy changes in the system are kinetic and gravitational.
	Expand the delta notation in terms of initial and final quantities.
	The initial gravitational energy is 0.
	Solve for the work done by the ground.
	Substitute the givens with units.
	Note the work done by the ground is positive as expected.

Example 4. A block of mass 1.8 kg is connected to a second block of mass 1.2 kg as shown in the diagram below. When the blocks are released from rest, they move through a distance of 0.30 m before the second block hits the floor. The coefficient of kinetic friction between block 1 and the table is 0.35. Find the speed of either block just before block 2 hits the floor. Ignore the mass of the pulley and the string, and assume that the string doesn't stretch.

Given: v_i = 0 m/s y_i,2 = 0.30 m y_f,2 = 0 m m₁ = 1.8 kg m₂ = 1.2 kg μ_k = 0.35 Goal: Find the speed v_f of either block before block 2 hits the floor. System: both blocks, Earth External forces: normal on block 1, tension on both blocks by the string initial state: both blocks at rest final state: both blocks moving at the same speed just before block 2 hits the floor Energy changes: ΔU_g is negative since block 2 decreases in elevation; ΔK is positive since both blocks speed up	The final state is shown by the dashed lines below. The direction of +y is defined to be up. The lowest position of the block 2 is designated as y_f= 0.
	.
	W_ext is composed of three terms. The normal force does 0 work on block 1, while kinetic friction does negative work. Together, the two tension forces do 0 work on the system. The tension force pulling to the right on block 1 does positive work, since the force is in the same direction as the displacement. The tension force pulling up on block 2 does negative work, since the force is opposite the displacement: W_T = Tdcos0° + Tdcos180° = 0.
	The energy changes in the system are kinetic and gravitational. We expand these in terms of the changes for each block. The kinetic energy changes can be grouped together, since the blocks move together. We keep the potential energy changes separate. Block 1 experiences no change in gravitational potential energy, since the elevation of that block doesn't change. Block 2 experiences a decrease in gravitational potential energy.
	The work done by kinetic friction on block 1 is W_f = μ_kNΔx(cosθ) = μ_kM₁gΔx(cos180°) = -μ_kM₁gΔx. Note that the horizontal displacement Δx of block 1 is the negative of the vertical displacement Δy = (y_2f - y_2i) of block 2. Since y_2f= 0, Δx = y_2i.
	We select the positive root since we are asked for speed.

Example 5. See this animation. At t = 0, a block attached to a spring is released from rest and oscillates horizontally about the origin on a table. There is friction between the block and the table. How much work does friction do on the block from t = 0 until the block returns for the first time to its turnaround point on the right? The mass of the block is 1.5 kg, and the spring constant of the spring is 460 N/m.

Given: x_i = 0.090 m x_f = 0.064 m v_i = 0 m/s v_f = 0 m/s m = 1.5 kg Goal: Find the work done by kinetic friction on the block from t = 0 to the first turnaround point on the right. System: block, spring External forces: normal, gravity, kinetic friction initial state: block at rest at x_i final state: block momentarily at rest at x_f Energy changes: ΔU_el is negative since the final extension of the spring is less than the initial extension. ΔK is 0.	x = 0 is the equilibrium position of the spring. The initial and final positions are read from the animation.
	.
	W_ext is composed of three terms. The normal and gravitational forces do 0 work on the block, while kinetic friction does negative work.
	The only energy change in the system is elastic. The initial and final velocities are 0, so the change in kinetic energy is 0. The elastic potential energy decreases.
	In finding the change in elastic potential energy, find the differences of the squares of the two extensions rather than finding the square of the difference. These two operations aren't equivalent.
	Note that it's not necessary to know the mass of the block to solve the problem.
	The work done by friction is negative as expected.