A 2.50-kg block on a horizontal floor is attached to a horizontal spring that is initially compressed 0.0300 m. The spring has force constant 840 N/m. The coefficient of kinetic friction between the floor and the block is µk = 0.40. The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of 0.0200 m from its initial position? (At this point the spring is compressed 0.0100 m.)

Solution 26E Step 1: Mass of block m = 2.50 kg Force constant k = 840 N/m Coefficient of kinetic friction k = 0.40 Horizontal spring that is initially compressed x = 0.0300 m Step 2: Energy stored in spring E = 1/2 kx2 E = 1/2 (840 N/m)(0.0300 m) 2 E = 0.378 J Friction force is given by F f energy stored in spring m*ss of block g * F f (0.378) (*.50 kg)(9.81 m/s ) 2 F = 9.270 N/m f Work done in friction W = energy stored in spring m*ss of block g bl* * moved distance 0.0200 m 2 W = 0.378 (*.81 m/s ) (0*0200 m) W = 0.1854 J Energy stored in spring when it has moved a distance of 0.020 m. 2 = 0.5 *40 N/m (0.*100 m) = 0.042