A box of mass m is released from rest at Point A , the top of a long, frictionless slide. Point

A box of mass m is released from rest at Point A , the top...

A box of mass $m$ is released from rest at Point $\mathrm{A}$ , the top of a long, frictionless slide. Point $\mathrm{A}$ is at height $H$ above the level of Points $\mathrm{B}$ and $\mathrm{C}$ . Although the slide is frictionless, the horizontal surface from Point $\mathrm{B}$ to $\mathrm{C}$ is not. The coefficient of kinetic friction between the box and this surface is $\mu_{k}$ , and the horizontal distance between Point $\mathrm{B}$ and $\mathrm{C}$ is $x .$ (a) Find the speed of the box when its height above Point $\mathrm{B}$ is $\frac{1}{2} H .$ (b) Find the speed of the box when it reaches Point $\mathrm{B}$ . (c) Determine the value of $\mu_{k}$ so that the box comes to rest at Point $\mathrm{C}$ . (d) Now assume that Points $B$ and $C$ were not on the same horizontal level. In particular, assume that the surface from B to C had a uniform upward slope so that Point C were still at a horizontal distance of $x$ from B but now at a vertical height of $y$ above B. Answer the question posed in part(c). (e) If the slide were not frictionless, determine the work done by friction as the box moved from Point A to Point B if the speed of the box as it reached Point B were half the speed calculated in part (b).

A box of mass $m$ is released from rest at Point $\mathrm{A}$ , the top of a long, frictionless slide. Point $\mathrm{A}$ is at height $H$ above the level of Points $\mathrm{B}$ and $\mathrm{C}$ . Although the slide is frictionless, the horizontal surface from Point $\mathrm{B}$ to $\mathrm{C}$ is not. The coefficient of kinetic friction between the box and this surface is $\mu_{k}$ ,
and the horizontal distance between Point $\mathrm{B}$ and $\mathrm{C}$ is $x .$
(a) Find the speed of the box when its height above Point $\mathrm{B}$ is $\frac{1}{2} H .$
(b) Find the speed of the box when it reaches Point $\mathrm{B}$ .
(c) Determine the value of $\mu_{k}$ so that the box comes to rest at Point $\mathrm{C}$ .
(d) Now assume that Points $B$ and $C$ were not on the same horizontal level. In particular, assume that the surface from B to C had a uniform upward slope so that Point C were still at a horizontal distance of $x$ from B but now at a vertical height of $y$ above B. Answer the question posed in part(c).
(e) If the slide were not frictionless, determine the work done by friction as
the box moved from Point A to Point B if the speed of the box as it reached Point B were half the speed calculated in part (b).

Key Concepts

Coefficient of Kinetic Friction

The coefficient of kinetic friction is a dimensionless parameter that quantifies the frictional resistance between two surfaces in relative motion. It is used in calculating the friction force, which is obtained by multiplying this coefficient by the normal force, and it is pivotal in problems involving energy dissipation on surfaces.

Work Done by Friction

Friction is a non-conservative force that converts mechanical energy into thermal energy, thereby reducing the system's total mechanical energy. The work done by friction is the product of the friction force and the distance over which it acts, and it is essential when analyzing energy losses in moving systems.

Gravitational Potential Energy

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is directly proportional to the object's mass, the gravitational acceleration, and its height above a chosen reference level, making it a fundamental concept in problems involving vertical motion.

Conservation of Energy

This principle states that within a closed system, energy is neither created nor destroyed but only transformed from one form to another. In mechanics, it is widely applied to relate gravitational potential energy and kinetic energy, taking into account any work done by non-conservative forces such as friction.

Kinetic Energy

Kinetic energy is the energy of an object due to its motion and is calculated as one half the mass times the square of its velocity. It plays a crucial role in dynamics, particularly in energy conservation calculations where potential energy is converted into kinetic energy.

Transcript

00:01 So in this problem, we have an incline, and this is point a, this is point b, and point c is over here.

00:13 So suppose this angle is theta.

00:16 From the problem, we know that from a to b, there is no any friction.

00:20 And from b to c, the friction constant is mu -k.

00:27 So for part a, if, so we want to know that, the velocity of the object.

00:42 So initially the object is over here, right? so we want to know the object if the object is at the height, which is half times h above the ground, which is over here.

00:54 So the total height is h.

00:56 And at this point, the height is half h.

01:01 So we can utilize the energy conservation law because there is no any friction on this surface, on the incline.

01:06 Okay? so we have to half h.

01:10 Xmg equal half mv squared right so this gives us v equals square root g times h and uh in part b so we want to know the velocity of the object at point b so we can see that uh the total height is h so h times m g equal half m vb square right? so this gives us vb equal square root 2 gh and in part c so suppose suppose the distance from b to c is x okay and we will have the total initial energy which is mgh they it will be all transferred into the friction it will be all transferred to overcome the friction because we want to make sure that the object is at a rest at point c.

02:17 So we have the relation mgh equal m g times mu k.

02:23 So this gives us the friction in the path from b to c, right? so friction times the total is x.

02:29 So this gives us the energy that consumed by the friction.

02:34 So we obtained that to mu k equal h over x.

02:37 All right.

02:39 And in part d, so for part d, we have a different model.

02:44 So this is still the incline with no any friction.

02:49 But so this is a point a, point b.

02:52 But from point b to c, there is another height above.

02:56 So it's something like this...