Solving for Distance and Minimum Coefficient of Friction

[jti3066]

Homework Statement

A wooden block of mass 3.2 kg is sitting near the bottom of a plance inclined at an angle of 23 deg relative to the horizontal. The block is given an initial velocity of 5.0 m/s up the plane. The coefficient of kinetic friction between the block and the surface of the inclined plane is 0.35.

A) How far up the plane does the block travel?

B) What must the minimum value of the coefficient of static friction between the block and the plane be for the block to remain at rest at the highest point of its trip?

Homework Equations

F_net = ma

v_f^2 - v_o^2 = 2(a)(x_f - x_i)

The Attempt at a Solution

A) y-comp: N = mg (since no acceleration in y direction)

x-comp: umgsin() = ma

a = ugsin()

0 - v_o^2 = 2(ugsin())(x_f)

-(5)^2 = 2 * .35 * 9.81 *sin(23)(x_f)

x_f = 9.31 m

B) f_s >= mgsin() = 3.2 * 9.81 * sin(23) = 12.26

 

[alex3]

Try thinking in terms of relative to the slope. Perpendicular to the slope, there is the normal force of the block. Parallel to the slope, the block experiences two forces, one due to the gravitational force on it's mass, and the other due to friction. You can use the superposition principal to get an acceleration parallel to the slope.

 

[jti3066]

Ok...sorry

B) y-component: N= mgcos()

x-component: mgsin()-uN = ma

so

a = gsin() - ugcos()

Correct?

 

[alex3]

looking good :)

except you're missing one sign, both forces on the block will be down the slope, not just the frictional force.

 

[jti3066]

missing one sign? Not sure what you meen...

 

[alex3]

You've put the frictional force with a negative sign, implying it's going down the slope (i.e. against the initial direction of motion), so what about the gravitational force, what sign should that have?

 

[jti3066]

Ok...(-g)...so:

a = gsin() + ugcos()

 

[alex3]

Negative g is correct, but it will yield

$$a = -(g\sin{\theta} + \mu g \cos{\theta})$$

This is because you're original assumption of making the frictional force negative (against the initial velocity) was correct, but it was correct because you treat g as negative. Then, the gravitational force is negative for exactly the same reason.

 

[jti3066]

A) I solved for a and got: a = -6.99

0 - (5^2) = 2(-6.99)(x_f)

x_f = 1.788 m

B) f_s >= mgsin() + umgcos()

 

[alex3]

For the second question, consider every force acting on the block: it has just reached the top of its ascent, so it will have gravity acting on it. The frictional force will always be opposing the direction of motion, so will be *up* the slope, so you require that the frictional force is greater than or equal to the gravitational force.

 

[jti3066]

B) f_s >= umgcos() ..

f_s >= 0.35(3.2)(9.81)cos(23) = 10.11