Lecture 10: Work and Energy

Work of A Force (1 of 2) A force does work on a particle when the particle undergoes a displacement along the line of action of the force. Work is defined as the product of force and displacement components acting in the same direction. So, if the angle between the force and displacement vector is q, the increment of work dU done by the force is dU  E ds cos  By using the definition of the dot product and integrating, the total work can be written as r2 U1 2   F .dr r1 Work of A Force (2 of 2) If F is a function of position (a common case) this becomes S2 U12   F cos  ds S1 If both F and q are constant F  F , c this equation further simplifies to U1 2  Fc cos  ( s2  s1 ) Work is positive if the force and the movement are in the same direction. If they are opposing, then the work is negative. If the force and the displacement directions are perpendicular, the work is zero. Work of A Weight The work done by the gravitational force acting on a particle (or weight of an object) can be calculated by using y2 U1 2    W dv y1 U1 2  W ( y2  y1 )  Wv The work of a weight is the product of the magnitude of the particle’s weight and its vertical displacement. If Dy is upward, the work is negative since the weight force always acts downward. Work of A Spring Force When stretched, a linear elastic spring develops a force of magnitude Fs  ks, where k is the spring stiffness and s is the displacement from the unstretched position. The work of the spring force moving from position s to position s is s2 1 s2 2 U1 2   Fs ds   k s ds  0.5k ( s2 ) 2  0.5k ( s1 ) 2 s1 s1 If a particle is attached to the spring, the force Fs exerted on the particle is opposite to that exerted on the spring. Thus, the work done on the particle by the spring force will be negative or U1 2  [0.5k ( s2 ) 2  0.5k ( s1 ) 2 ]. Spring Forces It is important to note the following about spring forces 1. The equations above are for linear springs only! Recall that a linear spring develops a force according to F = ks (essentially the equation of a line). 2. The work of a spring is not just spring force times distance at some point, i.e., ksi si . Beware, this is a trap that students often fall into! 3. Always double check the sign of the spring work after calculating it. It is positive work if the force on the object by the spring and the movement are in the same direction. Principle Of Work And Energy (1 of 2) By integrating the equation of motion,  F t  ma  mv dv/ds , the principle of work and energy can be written as t U 1 2  0.5m(v2 ) 2 or T1   U 12  T2 is the work done by all the forces acting on the particle as it moves from point 1 to point 2. Work can be either a positive or negative scalar.  U12 T1 and T2 are the kinetic energies of the particle at the initial and final position, respectively. Thus, T1  0.5 m v1 2 and 2 T2  0.5 m v 2  . The kinetic energy is always a positive scalar (velocity is squared!). So, the particle’s initial kinetic energy plus the work done by all the forces acting on the particle as it moves from its initial to final position is equal to the particle’s final kinetic energy. Principle Of Work And Energy (2 of 2) Note that the principle of work and energy T1  value. U12  T2  is not a vector equation! Each term results in a scalar Both kinetic energy and work have the same units, that of energy! In the SI system, the unit for energy is called a joule (J), where 1 J  1 N  m. In the FPS system, units are ft  lb The principle of work and energy cannot be used, in general, to determine forces directed normal to the path, since these forces do no work. The principle of work and energy can also be applied to a system of particles by summing the kinetic energies of all particles in the system and the work due to all forces acting on the system. Work of Friction Caused By Sliding The case of a body sliding over a rough surface merits special consideration. Consider a block which is moving over a rough surface. If the applied force P just balances the resultant frictional force μk N, a constant velocity v would be maintained. The principle of work and energy would be applied as 0.5m(v ) 2  Ps  (µ k N ) s  0.5m(v) 2 This equation is satisfied if P  μ N. However, we know from k experience that friction generates heat, a form of energy that does not seem to be accounted for in this equation. It can be shown that the work term μk Ns represents both the external work of the friction force and the internal work that is converted into heat. Example (1 of 3) Given: When s  0.6 m, the spring is not stretched or compressed, and the 10 kg block, which is subjected to a. force of 100 N, has a speed of 5 m/s down the smooth plane. Find: The distance s when the block stops. Plan: Since this problem involves forces, velocity and displacement, apply the principle of work and energy to determine s. Example (2 of 3) Solution: Apply the principle of work and energy between position 1 1s1  0.6m  and position 2 s2 . Note that the normal force (N) does no work since it is always perpendicular to the displacement. T1   U 1 2  T2 There is work done by three different forces; 1) work of a the force F =100 N; U F  100( s2  s1 )  100( s2  0.6) 2) work of the block weight; U W  10(9.81)( s2  s1 ) sin 30  49.05( s2  0.6) 3) and, work of the spring force. U S  0.5(200)( s2  0.6) 2  100( s2  0.6) 2 Example (3 of 3) The work and energy equation will be T1   U1 2  T2 0.5(10)52  100( s2  0.6)  49.05( s2  0.6)  100( s2  0.6) 2  0  125  149.05( s2  0.6)  100( s2  0.6) 2  0 Solving for (s2 - 0.6), (s 2  0.6)  {149.05  (149.052  4  (100)  125)0.5 } / 2(100) Selecting the positive root, indicating a positive spring deflection, (s2  0.6)  2.09m Therefore, s 2  2.69m