In Fig. 8-55, a block slides along a path that is without...

In Fig. 8-55, a block slides along a path that is without friction until the block reaches the section of length $L=0.75 \mathrm{~m}$, which begins at height $h=2.0 \mathrm{~m}$ on a ramp of angle $\theta=30^{\circ} .$ In that section, the coefficient of kinetic friction is $0.40 .$ The block passes through point $A$ with a speed of $8.0 \mathrm{~m} / \mathrm{s}$. If the block can reach point $B$ (where the friction ends), what is its speed there, and if it cannot, what is its greatest height above $A$ ?

Key Concepts

Gravitational Potential Energy

Gravitational potential energy represents the energy an object has due to its position in a gravitational field. When an object moves up or down an inclined plane, its potential energy changes accordingly, making it a key component in the energy balance when analyzing motions involving changes in height.

Kinetic Friction

Kinetic friction is the resistive force that opposes the motion of an object sliding across a surface. The frictional force is calculated using the coefficient of kinetic friction and the normal force. On an inclined plane, the normal force is modified by the cosine of the incline angle, which directly influences the amount of energy lost due to friction.

Conservation of Energy

This concept addresses the idea that the total mechanical energy in a system (in the absence of nonconservative forces) remains constant. In problems like this, energy conservation is applied by considering kinetic energy, gravitational potential energy, and the energy lost due to friction, allowing one to relate the initial energy at one point to that at another point along the path.

Work-Energy Theorem

The work-energy theorem states that the net work done on an object is equal to its change in kinetic energy. In scenarios where friction and other nonconservative forces are acting, this theorem helps in determining how the work done by these forces (like friction) affects the speed and height of an object along its path.

Transcript

00:01 So here we're going to refer to the point where it first encounters the rough region as point c.

00:06 So point c first encounters rough region.

00:19 We can say that here, this is the point at a height h above the reference level.

00:25 So we can then say using equation 817, we can find the speed at point c.

00:31 V .c would have to be equal to the square root of v .a squared, and then minus 2gh, this would be equal to the square root of 8 .0 squared, minus 2 times 9 .8 times a height of 2 .0.

00:53 And we find that v .c is going to equal approximately 4 .98 meters per second.

01:01 Now we need to see that its kinetic energy right at the beginning of its rough side.

01:07 So we can then say essentially k, the kinetic energy at point c, would be equal to 1 half m times 4 .98 quantity squared.

01:21 This is equaling 12 .4 approximately 12 .4 times m.

01:27 We are going to now carry along the mass and it will cancel out when we need.

01:34 Use equation 837.

01:39 We know that here the force normal is going to be equal to mg cosine of theta and we also know that here y is going to equal d sine of theta and we also note that if d is going to be less than l the block does not reach point b and this kinetic which this kinetic energy rather will turn into thermal energy the entire kinetic energy will turn into thermal and potential energy.

02:11 So we can say k c would be equal to m g y plus the force of kinetic friction times d.

02:21 Therefore we can say 12 .4m will equal m g d sign of theta plus the coefficient of kinetic friction m g d cosine of theta.

02:36 We can cancel out the ms now and we know that the coefficient of kinetic friction is equaling 0 .40 and we know that our theta is equaling 30 degrees...

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