A block is at rest on a plank whose angle can be varied. The angle is gradually increased from 0 deg. At 30.1°, the block starts to slide down the incline, traveling 3.40 m down the incline in 2.30 s. Calculate the coefficient of static friction between the block and the plank. Motion of a block Submit Answer Tries 0/12 Calculate the kinetic coefficent of friction between the block and the plank.

u_s = 0.5797

u_k = 0.4282

Explanation:

Given:

- The minimum angle Q = 30.1 degrees

- Distance traveled down the slope S (2.3) = 3.40 m

- Total time taken t = 2.30 s

Find:

a) coefficient of static friction between the block and the plank

b) kinetic coefficent of friction between the block and the plank.

Solution:

- Construct a FBD of the block. Take coordinate axis down the slope and normal to the slope. There three forces acting on the block at time. Weight components, Frictional force, and normal force.

- We are given the minimum amount of angle required to make the block move. Just enough to equal the static friction of the block. Hence we apply Newton's Law for static equilibrium as follows down the slope:

m*g*sin (Q) - F_f, s = 0

Where, F_f, s is the static friction.

- The static friction is proportional to the normal force acting on the body. Similarly apply Newton's Law for static equilibrium as follows normal to the slope:

m*g*cos (Q) - N = 0

N = m*g*cos (Q)

Where, N is the normal contact force acting on the body.

F_f, s = u_s*N

Where, u_s is the coefficient of static friction. Hence, the frictional force is:

F_f, s = u_s*m*g*cos (Q)

- Now we can calculate the coefficient of static friction from the down the slope equilibrium equation:

m*g*sin (Q) = u_s*m*g*cos (Q)

Simplify,

sin (Q) / cos (Q) = tan (Q) = u_s

- Plug in the angle Q:

u_s = tan (30.1) = 0.5797

- To determine the coefficient of kinetic friction we will consider the motion of block down the slope. Where the block travels a distance S (2.3) = 3.4 m.

- Assuming constant acceleration of the block, we can calculate the acceleration of the block by using second equation of motions down the slope as follows:

S (t) = S_o + V_o*t + 0.5a*t^2

- Where, S_o initial distance is 0, and V_o is also zero because the block was initially at rest Hence, we have:

S (t) = 0.5a*t^2

- Plug in the given values:

S (2.3) = 0.5*a * (2.3) ^2 = 3.40

a = 3.40*2 / (2.3) ^2

a = 1.2854 m/s^2

- Now we will apply the Newton's second law of motion down the slope for the block follows:

m*g*sin (Q) - F_f, k = m*a

Where, F_k, s is the kinetic friction.

- The kinetic friction is proportional to the normal force acting on the body. Similarly apply Newton's Law for static equilibrium as follows normal to the slope:

m*g*cos (Q) - N = 0

N = m*g*cos (Q)

Where, N is the normal contact force acting on the body.

F_f, k = u_k*N

Where, u_k is the coefficient of kinetic friction. Hence, the frictional force is:

F_f, k = u_k*m*g*cos (Q)

- Now we can calculate the coefficient of kinetic friction from the down the slope dynamic equation:

m*g*sin (Q) - u_k*m*g*cos (Q) = m*a

u_k = (g*sin (Q) - a) / g*cos (Q)

- plug in the values:

u_k = (9.81*sin (30.1) - 1.2854) / 9.81*cos (30.1)

u_k = 0.4282