Defining the kinetic energy, as 1 re have that V12=T2-71 or T1+w12=T2 (5) The above relationship is known as the principle of work and energg, and states that the mechanical work done on a particle is equal to the change in the kinetic energy of the particle Example Block sliding down an incline a block is released from rest at the top of a ramp. The coefficient of kinetic friction between the surface of the ramp and the block is u. We want to determine the velocity of the block as a function of the distance traveled on the ramp, s N 2 nng The forces on the block are: the weight, mg, the normal force, N, and, the friction force, uN. We have that f an and Ft= mat. Since Fn =N-mg cos a and an =0, we have n= mg cos a. Thus Ft= mg sina-HN= mg sin a- umg cos a, which is constant. If we apply the principle of work and energy between the position (1), when the block is at rest at the top of the ramp, and the position(), when the block has travelled a distance s, we have T1=0, T2=(mu)/2, and the work done by Ft is simply W12= fts. Thus T1+W12=T2 mg(sin- ucos a)=mu From which we obtain, for the velocit 2g(sin a-u cos a)s We make two observations: first, the normal force, N, does no work since it is, at all times, perpendicular to the path, and second, we have obtained the velocity of the block directly without having to carry out ny integrations. Note that an alternative, longer approach would have been to directly use F= ma, and integrate the corresponding expression for the acceleration. The use of T to denote the kinetic energy, instead of K, is customary in dynamics textbooksDefining the kinetic energy1 , as T = 1 2 mv2 , we have that, W12 = T2 − T1 or T1 + W12 = T2 . (5) The above relationship is known as the principle of work and energy, and states that the mechanical work done on a particle is equal to the change in the kinetic energy of the particle. Example Block sliding down an incline A block is released from rest at the top of a ramp. The coefficient of kinetic friction between the surface of the ramp and the block is µ. We want to determine the velocity of the block as a function of the distance traveled on the ramp, s. The forces on the block are: the weight, mg, the normal force, N, and, the friction force, µN. We have that Fn = man and Ft = mat. Since Fn = N − mg cos α and an = 0, we have N = mg cos α. Thus, Ft = mg sinα − µN = mg sin α − µmg cos α, which is constant. If we apply the principle of work and energy between the position (1), when the block is at rest at the top of the ramp, and the position (2), when the block has travelled a distance s, we have T1 = 0, T2 = (mv2 )/2, and the work done by Ft is simply W12 = Fts. Thus, T1 + W12 = T2 , or, mg(sin α − µ cos α) = 1 2 mv2 . From which we obtain, for the velocity, v = p 2g(sinα − µ cos α)s . We make two observations: first, the normal force, N, does no work since it is, at all times, perpendicular to the path, and second, we have obtained the velocity of the block directly without having to carry out any integrations. Note that an alternative, longer approach would have been to directly use F = ma, and integrate the corresponding expression for the acceleration. 1The use of T to denote the kinetic energy, instead of K, is customary in dynamics textbooks 3