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Illustration 6.2: Constant Forces




|F applied| = N | m = kg | μk =

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This Illustration shows a block on a table subjected to an applied force and a frictional force. The mass of the block, the force applied, and the coefficient of kinetic friction can be controlled by typing values for these variables into the text boxes and pressing the "set values and play" button. The coefficient of static friction is fixed at μs = 0.4. Restart.

Consider the animation with a 100-kg block and vary the force applied from 0 to 391 N. What happens in the animation? The block does not move. What is the work done on the block due to the applied force? What is the work done on the block due to the force of static friction? What is the work done on the block due to the normal force? What is the work done on the block due to the force of gravity? The work done by each of these forces is zero. How do you know? First there is no displacement. If there is no displacement there can be no work. In addition, the normal force and the force of static friction can never do work. The normal force cannot do work because if there is a displacement it would always be perpendicular to the normal force, hence no work. The force of static friction can also never do work. When there is static friction there can never be a displacement. After all, static friction implies the block is static and thus not moving. Since there is no work done on the block, the block's kinetic energy cannot change.

Now consider the animation with a 100-kg block and an applied force of 446 N. What happens in the animation? The block moves and, in fact, it accelerates. What is the work done on the block due to the applied force? What is the work done by the force of kinetic friction? What is the work done on the block due to the normal force? What is the work done on the block due to the force of gravity? The net work done by the sum of these forces is now not zero. How do you know? There is a change in kinetic energy. This can only happen when there is work done on the block. The force of gravity does not do any work on the block because the force is perpendicular to the displacement in this animation. The normal force, as said above, can never do work. The work done by the applied force will be positive.

The force of kinetic friction reduces the kinetic energy of the block by | Fk friction Δx | because the frictional force and the block's displacement are in opposite directions. Kinetic friction will always oppose motion, so it will always reduce kinetic energy. Note that we do not say the work done on the block by kinetic friction. This phrase is not correct. The work done by friction on the block is the energy the block loses, and this is not equal to - Fk friction Δx. Some of the kinetic energy dissipated by friction, | Fk friction Δx |, is transferred to the table as thermal energy (the table heats up), while some of it remains with the block as thermal energy (the block heats up). Therefore, -Fk friction Δx is not the work done on the block by friction: - Fk friction Δx is the total work done by friction (done on the block and the table) and is the amount that the kinetic energy is reduced.

The change in kinetic energy of the block is the net force Fapplied - Fk friction times the displacement (this is because the net force points in the direction of the displacement). Therefore, in the table, Fnet times x = KE (because x is the displacement since the block starts at x = 0 m and the final KE is the change in KE since the block starts with no kinetic energy).

Illustration authored by Mario Belloni.
Script authored by Steve Mellema and Chuck Niederriter and modified by Mario Belloni.

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