How Does Kinetic Friction Affect Block Acceleration?

[chocolatelover]

Homework Statement

A 425 g block is at rest on a horizontal surface. The coefficient of static friction between the block and the surface is .770, and the coefficient of kinetic friction is .220. A force of magnitude P pushes the block forward and downward. Assume the force is aplied at an angle of 40°

a. Assuming P is large enought to made the block move, find the acceleration of the block as a function of P (Use P as necessary and round numbers to the third decimal place)

b. If P=11.13 N and then P=22.26, find the accelerations and the frictional forces exerted on the block

c. What is the minimum acceleration for the block when P is a minimum and motion is just ready to begin?

Homework Equations

The Attempt at a Solution

a. ΣFx=Pcosθ -f=ma
ΣFxy=N-mg-Psinθ=0

f=μN

Pcosθ-μΝ=ma

Pcosθ-μ(mg+sinθ)=ma

(cosθ/m)P-mg+(μsinθ/m)P

-mg+μsinθ/m +(cosθ/m)P=a

(-425g)(9.8m/s^2)+(.220sin40/425g)+(cos40/424)P=a

a=4165.001+.001P

b. P=11.13

a=4165.001+.001(11.13N)
4165.012

f=μΝ
N=(4165.012)(11.13)+425g(9.8m/s^2)

c. For part c. I was thinking about using Calculus to minimize the equation, would that work?

Could someone please tell me if this is correct so far and if it isn't could you please show me what to do?

Thank you very much

 

[chocolatelover]

Could someone please tell me if this looks correct?

a.
μs=.770
μk=.220

ΣFx=max

ΣFx=Pcos40-f
f=μkN
may=0
0=Σfy=N-mg-Py
=N-mg-Psin40
=Pcos40-[.220(mg+Psin40)]
ΣFx=max
Pcos40-.220(mg+psin40)=max
pcos40/.425kg-.220(.425(9.8)+psin40)/.425=ax
1.47p-2.155=ax

b.P=11.13N
1.47(11.13N)-2.155=ax
ax=14.206m/s^2

f=μkN

=.220(9.8m/s^2)
f=2.156

c. I would then do the same thing for part c, right?

Thank you very much

 

[Moetasim]

for your help.


I would say that your approach to solving the problem is correct. You have correctly applied the equations for Newton's second law and the forces acting on the block. Your calculations for part a and b are also correct.

For part c, using calculus to minimize the equation is a valid approach. You can also use the fact that at the point where the block is just about to move, the net force acting on it is zero. This means that the force applied (P) is equal to the maximum static friction force (μN). So you can set P equal to μN and solve for the minimum acceleration.

Overall, your solution is well-thought-out and correct. Great job!