Question

Two boxes with masses m1=5.0 kg and m2=3.0 kg are connected by a massless rope, which is put over a massless, frictionless pulley. The coefficient of kinetic friction between the incline and the boxes is 0.12 and the coefficient of static friction between the incline and the boxes is 0.25.The angles of the inclines with respect to the horizontal are ?=45** and ?2=30**. Use 9.8 m/s^2 for gravity and assume the incline is fixed in space.

a.) Show that the system will have a nonzero acceleration if it is initially at rest. Find the magnitude and direction of acceleration.

b.) If the system moves 1.5 m(in the direction found in part a), what is the work done by: i.) gravity on m1, ii.) the normal force on m1, iii.) friction on m1, iv.) gravity on m2, v.) the normal force on m2, vi.) friction on m2

c.) Using the results from part b, what is the speed of the system after it has moved 1.5 m if it was initially at rest?

Answer 1

the diagram for the problem looks like this

First we draw the Force diagram for each Box m₁ and m₂

Force Diagram for Box m1 :

m₁ = mass of box = 5 kg , g = 9.8 m/s²   Fn1 = normal force by incline

m₁ g = force of gravity = m₁ g = (5 kg) (9.8 m/s² ) = 49 N

m₁ g Cos45 = Component of force of gravity perpendicular to incline = (5 kg) (9.8 m/s² ) Cos45 = 34.65 N

m₁ g Sin45 = Component of force of gravity Parallel to incline = (5 kg) (9.8 m/s² ) Sin45 = 34.65 N

F_n1 = normal force by incline

T = tension force in rope

f₁= force of friction which is given by the formula

f₁ = u_s F_n1 (when system at rest)

f1 = u_k F_n1 (when system in motion)

Now we write the Force equation for Box m₁

Force equation Perpendicular to incline : F_n1 - m₁g cos45 =0

F_n1 = m₁ gcos45 = 34.65 N

Force equation Parallel to incline: T - m1g sin45 - f1 =ma

frictional force f₁ is given as

f₁ = u_s F_n1   (when system at rest)

f₁ = (0.25)(34.65 N) = 8.66 N (since u_s = 0.25 and F_n1= 34.65)

inserting the value of f1 in the equation we get

T - 34.65 - 8.66 =5a (since m₁ = 5 kg)

T - 43.31 = 5a Eq-1

Force Diagram for Box m₂ :

m₂ = 3 kg ,

Force equation for Box m₂ :

Force equation Perpendicular to incline : F_n2 - m₂g cos45 =0

F_n2 = m₂ gcos30 = 25.46 N

Force equation Parallel to incline: m₂g sin30 - f₂ - T=ma

frictional force f₂ is given as

f₂ = u_s F_n2 (when system at rest)

f₂ = (0.25)(25.46 N) = 6.37 N (since u_s = 0.25 and F_n2 = 25.46 N)

inserting the value of f₂ in the equation we get

14.7 - 6.37 - T =3a (since m₁ = 5 kg)

8.33 - T = 3a Eq-2

Now we solve the two equations

T - 43.31 = 5a Eq-1

8.33 - T = 3a Eq-2

Adding Eq-1 and Eq-2

T - 43.31 + 8.33 - T = 5a + 3a

a = - 4.37 m/s²

Since the acceleration is negative , the direction of motion is in opposite direction of what we have assumed. that is , Box m₁ goes down parallel to incline and Box m₂ goes up parallel to incline.

b)

i) distance parallel to incline = 1.5 m

Force =F= m₁ g Sin45 = Component of force of gravity Parallel to incline = (5 kg) (9.8 m/s² ) Sin45 = 34.65 N

= angle between F and d = 0 ( since F is in same direction as the distance)

W = F d cos = (34.65 N)(1.5 m) Cos0 = 51.98 J

ii) distance parallel to incline = 1.5 m

Force =F= F_n1 = Normal force perpendicular to incline = 34.65 N

= angle between F and d = 90 ( since F is in perpendicular direction as the distance)

W = F d cos = (34.65 N)(1.5 m) Cos90 = 0 J

iii) distance parallel to incline = 1.5 m

Force =F= f₁ = frictional force on m₁ which is given as

f₁ = u_k F_n1   (when system in motion)

f₁ = (0.12)(34.65 N) = 4.16 N (since u_k = 0.12 and F_n1 = 34.65)

= angle between F and d = 180 ( since F is in opposite direction as the distance)

W = F d cos = (4.16 N)(1.5 m) Cos180 = -6.24 J

iv)  distance parallel to incline = 1.5 m

Force =F= m₂ g Sin30 = Component of force of gravity Parallel to incline = (3 kg) (9.8 m/s² ) Sin30 = 14.7 N

= angle between F and d = 180 ( since F is in Opposite direction as the distance)

W = F d cos = (14.7 N)(1.5 m) Cos180 = -22.05 J

v) distance parallel to incline = 1.5 m

Force =F= F_n2 = Normal force perpendicular to incline = 25.46 N

= angle between F and d = 90 ( since F is in perpendicular direction as the distance)

W = F d cos = (34.65 N)(1.5 m) Cos90 = 0 J

vi) distance parallel to incline = 1.5 m

Force =F= f₂ = frictional force on m₂ which is given as

f₂ = u_k F_n2 (when system in motion)

f2 = (0.12)(25.46 N) = 3.06 N (since u_k = 0.12 and F_n2 = 25.46)

= angle between F and d = 180 ( since F is in opposite direction as the distance)

W = F d cos = (3.06 N)(1.5 m) Cos180 = -4.59 J

c) initial velocity of the system = V_i =0

acceleration = a = 4.37 m/s^₂

d = 1.5 m

using the equation , Vf^2 = Vi^2 + 2 ad

Vf^2 = (0)^2 + 2 (4.37)(1.5)

Vf = 3.62 m/s