Help please! Help please! ) SptS mi A block with mass m, is placed on an incline with angle ?. m2 The mass m, is connected to a second mass m2 by placing a cord over a frictionless pulley (which can be considered massless). The coefficient of kinetic friction is Hk a) Find the mass m2 for which the block m, moves up the plane at constant speed V. constant speed v once set into motion. if they are released at rest. b) Find the mass m2 for which block m, moves down the plane at c) Over what range of values of m2 will the blocks remain at rest Solution part a: if it is moving at constant speed, then acceleration of the system is 0. let tension in the cord be T. friction force is acting along the downward direction of the incline. friction force=friction coefficient*normal force =mu_k*m1*g*cos(alpha) balancing forces for m2: T=m2*g…(1) balancing forces for m1: T-weight along the incline-friction force=0 ==>T=m1*g*sin(alpha)+mu_k*m1*g*cos(alpha) ==>m2*g=m1*g*(sin(alpha)+mu_k*cos(alpha)) ==>m2=m1*(sin(alpha)+mu_k*cos(alpha)) part b: if it is moving at constant speed, then acceleration of the system is 0. let tension in the cord be T. friction force is acting along the upward direction of the incline. friction force=friction coefficient*normal force =mu_k*m1*g*cos(alpha) balancing forces for m2: T=m2*g…(1) balancing forces for m1: weight along the incline-friction force-T=0 ==>T=m1*g*sin(alpha)-mu_k*m1*g*cos(alpha) ==>m2*g=m1*g*(sin(alpha)-mu_k*cos(alpha)) ==>m2=m1*(sin(alpha)-mu_k*cos(alpha)) part c: if they are released at rest, over the range of m1*(sin(alpha)-mu_k*cos(alpha)) and m1*(sin(alpha)+mu_k*cos(alpha)), the block will be at rest. .

1. Help please!
Help please!
) SptS mi A block with mass m, is placed on an incline with angle ?. m2 The mass m, is
connected to a second mass m2 by placing a cord over a frictionless pulley (which can be
considered massless). The coefficient of kinetic friction is Hk a) Find the mass m2 for which the
block m, moves up the plane at constant speed V. constant speed v once set into motion. if they
are released at rest. b) Find the mass m2 for which block m, moves down the plane at c) Over
what range of values of m2 will the blocks remain at rest
Solution
part a:
if it is moving at constant speed, then acceleration of the system is 0.
let tension in the cord be T.
friction force is acting along the downward direction of the incline.
friction force=friction coefficient*normal force
=mu_k*m1*g*cos(alpha)
balancing forces for m2:
T=m2*g…(1)
balancing forces for m1:
T-weight along the incline-friction force=0
==>T=m1*g*sin(alpha)+mu_k*m1*g*cos(alpha)
==>m2*g=m1*g*(sin(alpha)+mu_k*cos(alpha))

2. ==>m2=m1*(sin(alpha)+mu_k*cos(alpha))
part b:
if it is moving at constant speed, then acceleration of the system is 0.
let tension in the cord be T.
friction force is acting along the upward direction of the incline.
friction force=friction coefficient*normal force
=mu_k*m1*g*cos(alpha)
balancing forces for m2:
T=m2*g…(1)
balancing forces for m1:
weight along the incline-friction force-T=0
==>T=m1*g*sin(alpha)-mu_k*m1*g*cos(alpha)
==>m2*g=m1*g*(sin(alpha)-mu_k*cos(alpha))
==>m2=m1*(sin(alpha)-mu_k*cos(alpha))
part c:
if they are released at rest, over the range of m1*(sin(alpha)-mu_k*cos(alpha)) and
m1*(sin(alpha)+mu_k*cos(alpha)), the block will be at rest.