Question

QUESTION 5

A mass of 12 kg is placed on a horizontal surface with friction with a coefficient of static friction of 0.81 and coefficient of kinetic friction of 0.44. A force of 13 Newtons is applied in the -y direction and a force of 16 Newtons is also applied in the -y direction. Another force is applied in the +x direction (see sketch). The force applied in the +x direction is increased until it is 1.9 times the force needed to overcome static friction. If the mass is initially at rest, how far does the mass move after these forces are applied for 3.2 seconds? Answer in meters.

QUESTION 6

A mass of 16 kg is placed on a horizontal surface with friction with a coefficient of static friction of 0.86 and coefficient of kinetic friction of 0.22. A force of 28 Newtons is applied in the -y direction and a force of 5 Newtons is applied in the +y direction. Another force is applied in the +x direction (see sketch). The force applied in the +x direction is increased until it is 2.3 times the force needed to overcome static friction. If the mass is initially at rest, what is the magnitude of the object's final velocity, in m/s, after it has moved 5 meters in the +x direction after the forces are applied?

Answer 1

5.

IN vertical,

Fy_net = N - mg - 13 - 16 = 0

N = 29 + (12 x 9.8) = 146.6 N

Static friction, fs = us N = 0.81 x 146.6 = 118.75 N

so F = 1.9 Fs = 225.6 N

kinetic friction, fk = uk N = 64.5 N
In horizontal,

Fnet = m a

F - fk = m a

225.6 - 64.5 = 12a

a = 13.4 m/s^2

d = u t + a t^2 /2 = 0 + (13.4 x 3.2^2 / 2) = 68.7 m .....Ans

6. N vertical,

Fy_net = N - mg - 28 + 5 = 0

N = 179.8 N

Static friction, fs = us N = 0.86 x 146.6 = 154.6 N

so F = 2.3 Fs = 355.6 N

kinetic friction, fk = uk N = 39.5 N
In horizontal,

Fnet = m a

F - fk = m a

a = 19.75 m/s^2

v^2 - u^2 = 2 a d

v^2 - 0^2 = 2(19.75)(5)

v = 14 m/s

Answer 2

5.

IN vertical,

Fy_net = N - mg - 13 - 16 = 0

N = 29 + (12 x 9.8) = 146.6 N

Static friction, fs = us N = 0.81 x 146.6 = 118.75 N

so F = 1.9 Fs = 225.6 N

kinetic friction, fk = uk N = 64.5 N
In horizontal,

Fnet = m a

F - fk = m a

225.6 - 64.5 = 12a

a = 13.4 m/s^2

d = u t + a t^2 /2 = 0 + (13.4 x 3.2^2 / 2) = 68.7 m .....Ans

6. N vertical,

Fy_net = N - mg - 28 + 5 = 0

N = 179.8 N

Static friction, fs = us N = 0.86 x 146.6 = 154.6 N

so F = 2.3 Fs = 355.6 N

kinetic friction, fk = uk N = 39.5 N
In horizontal,

Fnet = m a

F - fk = m a

a = 19.75 m/s^2

v^2 - u^2 = 2 a d

v^2 - 0^2 = 2(19.75)(5)

v = 14 m/s