Part A Learning Goal: To apply the principle of linear impulse and momentum to a
ID: 2086787 • Letter: P
Question
Part A Learning Goal: To apply the principle of linear impulse and momentum to a system of particles. Integrating the equation of motion, as applied to all particles in a system, yields Sm; (vi) + . F?dt= m; (vi) where m, is the ith particle's mass, v, is the ith particle's velocity, and F, is the external force that acts on the ith particle. This relationship states that the sum of the initial linear momenta, at time ti, and the impulses of all the external forces acting between times t1 and t2 is equal to the sum of the linear momenta of the system, at time to. If the system has a mass center, G, the expression becomes Two blocks, each of mass m = 5.10 kg, are connected by a massless rope and start sliding down a slope of incline 8 = 36.0° at t = 0.000 s. The slope's top portion is a rough surface whose coefficient of kinetic friction is r = 0.260. At a distance d = 2.10 m from block A's initial position the slope becomes frictionless. (Figure 1) What is the velocity of the blocks when block A reaches this frictional transition point? Assume that the blocks' width is negligible. Express your answer numerically in meters per second to four significant figures. ? View Available Hint(s) % AZ¢ vecs ? m/s m(VG) + E, F, dt=m(VG)2 This expression allows the principle of linear impulse and momentum to be applied to a system of particles that is represented as a single particle. Submit Figure Part B This question will be shown after you complete previous question(s). Part C This question will be shown after you complete previous question(s).Explanation / Answer
The mass of the each block is m = 5.10 kg
The coefficient of friction is ? = 0.26
The distance of the frictional surface is d = 2.10 m
The angle of inclination is ? = 36 degree
The forces on the incline plane are
mgsin? = 5.10*9.8*sin36 = 29.40 N
component of weight along the incline
?mgcos? = 0.26*5.10*9.8*cos36 = 10.52 N
The net force along the incline is
F = 29.40 - 10.52 = 18.88 N
so the acceleration is
a = F/m
a = 18.8/5.10
a = 3.70 m/s^2
The time to travel 2.10 m is
s = 0.5at2
t = ?(2*2.10/3.70)
t = 1.0654 s;
v = at
v = 3.70*1.0654
v = 3.9420 m/s