CALC On a winter day in Maine, a warehouse worker is shoving...

CALC On a winter day in Maine, a warehouse worker is shoving boxes up a rough plank inclined at an angle $\alpha$ above the horizontal. The plank is partially covered with ice, with more ice near the bottom of the plank than near the top, so that the coefficient of friction increases with the distance $x$ along the plank: $\mu=A x,$ where $A$ is a positive constant and the bottom of the plank is at $x=0 .$ (For this plank the coefficients of kinetic and static friction are equal: $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu .$ The worker shoves a box up the plank so that it leaves the bottom of the plank mov- ing at speed $v_{0}$ . Show that when the box first comes to rest, it will remain at rest if $$v_{0}^{2} \geq \frac{3 g \sin ^{2} \alpha}{A \cos \alpha}$$

CALC On a winter day in Maine, a warehouse worker
is shoving boxes up a rough plank inclined at an angle $\alpha$ above
the horizontal. The plank is partially covered with ice, with
more ice near the bottom of the plank than near the top, so that
the coefficient of friction increases with the distance $x$ along the
plank: $\mu=A x,$ where $A$ is a positive constant and the bottom of
the plank is at $x=0 .$ (For this plank the coefficients of kinetic
and static friction are equal: $\mu_{\mathrm{k}}=\mu_{\mathrm{s}}=\mu .$ The worker shoves
a box up the plank so that it leaves the bottom of the plank mov-
ing at speed $v_{0}$ . Show that when the box first comes to rest, it
will remain at rest if
$$v_{0}^{2} \geq \frac{3 g \sin ^{2} \alpha}{A \cos \alpha}$$

Key Concepts

Newton's Laws of Motion

Newton's laws, especially the second law, are fundamental to analyzing the forces acting on the box along the inclined plane. They provide the relationship between the net force on the box (including gravitational and frictional components) and its acceleration. This framework allows us to understand how the box decelerates as it moves up the incline and eventually comes to rest.

Work-Energy Principle

The work-energy principle connects the work done by both the gravitational force and friction with the change in the kinetic energy of the box. In this problem, it is used to relate the initial kinetic energy provided by the box's speed to the energy dissipated by work done against gravity and friction as the box ascends the plane.

Friction Force and Friction Coefficient

Friction, which opposes the motion of the box, is characterized by a coefficient that may vary along the path. Understanding both kinetic and static friction is crucial in determining whether the box will remain at rest once it stops. In cases with a variable friction coefficient, the frictional force becomes a function of position, adding complexity to the analysis.

Integration in Mechanics

With a friction coefficient that increases with distance along the plane, calculus—specifically integration—is required to accumulate the effects of the frictional force over the distance traveled by the box. This technique allows for the calculation of the total work done against friction in a system where friction is not constant, thereby contributing to a more accurate energy balance.

Transcript

00:01 In maine on a winter's day, a warehouse worker is shoving boxes up a rough plank.

00:05 The plank is inclined at an angle above the horizontal, and that angle is this.

00:11 The plank is partially covered with ice, more ice near the bottom.

00:16 So the coefficient of friction increases with distance x on the plank.

00:21 If the coefficient of friction equals ax, where a is a positive constant, and the bottom of the plank is x equals zero, zero then we know that we're told that the coefficient of static friction the coefficient coefficient of kinetic friction is equal just to coefficient of friction so the worker pushes a box at the plank so it leaves the bottom of the plank moving with a speed v o we're asked to show when a box first comes to rest it will remain at rest if this equation is true let's begin again.

00:52 Okay.

00:53 So this is going to be pushing my math skills to the absolute end of time.

00:59 So our w will be equal to and dx will equal one half and bf minus bo squared.

01:18 Squared.

01:20 Okay.

01:21 Then the force, the forces acting on the box will be our friction minus our mg times the sine of theta.

01:39 And that will equal negative ax mg times the cosine of theta minus minus mg times the sine.

02:05 That's not theta.

02:08 Okay, then substituting we'll get and then we'll take that minus mg times our sine dx and this will equal negative a squared mg cosine over 2 minus minus m g x sine.

03:01 And then we'll combine our equations.

03:08 And let's call this equation three.

03:12 And let's call this one equation two...

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