Question
In the figure below, a small block is sent through point A with a speed of 6.4 m/s. Its path is without friction until it reaches the section of length L = 12 m, where the coefficient of kinetic friction is 0.70. The indicated heights are h1 = 5.2 m and h2 = 1.0 m. What is the speed of the block at point B? m/s What is the speed of the block at point C? m/s Does the block reach point D? If so, what is its speed there and if not, how far through the section of friction does it travel? in units of
Explanation / Answer
According to law of conservation ofenergy K1 + U1 = K2 + U2 (1/2) * m *v12 + m * g* h1 = (1/2) * m *v22 + m * g* h2 (1/2)* v12 + g *h1 = (1/2) *v22 + g *h2 a. (1/2) *vA2 + g *hA = (1/2) *vB2 + g *hB 0.5 *6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8* 0 vB = (71.44* 2) = 11.95 m/s b. (1/2)* vA2 + g *hA = (1/2) *vC2 + g *hC 0.5* 6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8 *1.0 vC = (71.44- 9.8) * 2 = 11.10 m/s c. Accordingto work energy theorem Workdone = loss in k.e. hencethe block will cross length L if the k.e. is more thanwork done. W = (* m * g) * L = 0.7* m * 9.8 * 12 = 82.32* m J k.e. K = (1/2)* m * vC2 = 0.5* m * 11.102 = 61.61* m J Since k.e. is less than work donerequired, the block will not reach D. Further K = Wl Wl is thework done is travelling distance l. 61.61* m = ( * m * g) *l distance l = 61.61/ 0.7 * 9.8 = 8.98 m a. (1/2) *vA2 + g *hA = (1/2) *vB2 + g *hB 0.5 *6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8* 0 vB = (71.44* 2) = 11.95 m/s 0.5 *6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8* 0 vB = (71.44* 2) = 11.95 m/s b. (1/2)* vA2 + g *hA = (1/2) *vC2 + g *hC 0.5* 6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8 *1.0 vC = (71.44- 9.8) * 2 = 11.10 m/s 0.5* 6.42 + 9.8 *5.2 = 0.5 *v22 + 9.8 *1.0 vC = (71.44- 9.8) * 2 = 11.10 m/s c. Accordingto work energy theorem Workdone = loss in k.e. hencethe block will cross length L if the k.e. is more thanwork done. W = (* m * g) * L = 0.7* m * 9.8 * 12 = 82.32* m J k.e. K = (1/2)* m * vC2 = 0.5* m * 11.102 = 61.61* m J Since k.e. is less than work donerequired, the block will not reach D. Further K = Wl Wl is thework done is travelling distance l. 61.61* m = ( * m * g) *l distance l = 61.61/ 0.7 * 9.8 = 8.98 m Wl is thework done is travelling distance l. 61.61* m = ( * m * g) *l distance l = 61.61/ 0.7 * 9.8 = 8.98 m