Question
Please answer the above questions (A), (B) & (C)
A block of mass 3.0 kg is attached to a horizontal spring that has a force constant of 1.20 times 103 N/m, and is free to slide on a frictionless surface as shown. The spring is compressed to xj = -7.5 cm by pushing on the block, and then the block is released. (A) Find the work done by initially compressing the spring. Find the kinetic energy of the block when it reaches x = 0. Find the speed of the block at x=0. No need to transcribe Find the initial total mechanical energy. The external force first does work on the system by pushing the block to compress the spring, whose potential energy increases by delta PE=1/2kxi2=1/2(1.20 times 103 N/m)(7.50 times 10-2 N/m)2= J. The work done on the system is this change in potential energy Delta PE. Find the kinetic energy at x = 0. The force from the spring does work on the block, increasing its kinetic energy and speed until it reaches x = 0. As this occurs, the reaction force of the block does negative work on the spring to decrease its potential energy. The total mechanical energy of the system: E mechanical = KE + PE = 1/2mv2 + 1/2kx2 remains constant throughout. At the moment of release, all of the work done in compressing the spring has been transformed into kinetic energy of the block. The kinetic energy Delta KE = 1/2mvi2 gained by the block is then the potential energy Delta PE = 1/2kxj2 of the spring. This was calculated in part (A) and is equal to J. Find the speed at x = 0. As the block moves, the mechanical energy of the system remains equal to the amount of work Delta U = 1/2kxj2 done in first compressing the spring. E mechanical = 1/2kxj2 At x = 0, all the mechanical energy is kinetic energy. E mechanical = 1/2mvf2 Substituting Equation (1) for the value of E mechanical into Equation (2) gives an expression for the speed at x = 0. vf2= xj2k/m vf=xj(k/m)1/2 = m/s
Explanation / Answer
1. work done = kx^2/2 = 1200 x 0.075^2/2 = 3.375 J
2. K.E> = kx^2/2 = 3.375 J
3. k.e. = mv^2/2
3.375 = 3 x v^2/2
v = 1.5 m/s