To find the expression for the speed of the object at the bottom of the slope, we will apply the concepts of energy conservation, accounting for the work done by friction. Here’s the step-by-step process:

Step 1: Gravitational potential energy at the top

The object has gravitational potential energy at the top of the slope. This energy is given by:

$PE_{\text{top}} = mgh$

where:

• $m$ is the mass of the object,
• $g$ is the acceleration due to gravity, and
• $h$ is the height of the slope.

Step 2: Kinetic energy at the bottom

At the bottom of the slope, the object has kinetic energy but no potential energy. The kinetic energy is:

$KE_{\text{bottom}} = \frac{1}{2}mv^2$

where $v$ is the speed of the object at the bottom of the slope (which is what we are solving for).

Step 3: Work done by friction

As the object slides down the slope, friction does work that opposes the motion. The work done by friction is:

$W_{\text{friction}} = F_{\text{friction}} \cdot d = \mu_k N \cdot L$

where:

• $\mu_k$ is the coefficient of kinetic friction,
• $N$ is the normal force on the object,
• $L$ is the length of the slope (distance traveled by the object),
• $F_{\text{friction}} = \mu_k N$, and
• $d = L$ (the length of the slope).

The normal force $N$ is the component of the object's weight perpendicular to the slope. If $\theta$ is the angle of the slope, then:

$N = mg \cos \theta$

Thus, the work done by friction becomes:

$W_{\text{friction}} = \mu_k mg \cos \theta \cdot L$

Step 4: Energy conservation with work done by friction

Using the principle of conservation of energy, the initial potential energy at the top is converted into kinetic energy at the bottom, minus the work done by friction:

$mgh = \frac{1}{2}mv^2 + \mu_k mg \cos \theta \cdot L$

Step 5: Relating $\cos \theta$ to known quantities

From the geometry of the slope, we know:

$\sin \theta = \frac{h}{L}$

Thus, $\cos \theta$ can be written as:

$\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left( \frac{h}{L} \right)^2}$

Step 6: Solve for $v$

Now, solve for $v$ by isolating $v^2$ on one side of the equation:

$mgh - \mu_k mg \cos \theta \cdot L = \frac{1}{2}mv^2$

$2mgh - 2\mu_k mg \cos \theta \cdot L = mv^2$

$v^2 = 2g \left( h - \mu_k L \cos \theta \right)$

Finally, substitute $\cos \theta = \sqrt{1 - \left( \frac{h}{L} \right)^2}$:

$v^2 = 2g \left( h - \mu_k L \sqrt{1 - \left( \frac{h}{L} \right)^2} \right)$

Taking the square root of both sides:

$v = \sqrt{2g \left( h - \mu_k L \sqrt{1 - \left( \frac{h}{L} \right)^2} \right)}$

This is the expression for the object's speed at the bottom of the slope in terms of $m$, $\mu_k$, $L$, $h$, and $g$.

Would you like more details on any part of the solution?

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Here are 5 related questions to deepen understanding:

1. How does increasing the coefficient of kinetic friction affect the object's final speed?
2. What happens to the object's final speed if the slope becomes steeper (i.e., $h$ increases)?
3. How would the expression change if the slope were not smooth (with much higher friction)?
4. Can you derive the final speed if there was no friction (i.e., $\mu_k = 0$)?
5. What is the physical significance of the angle $\theta$ in this problem?

Tip: In problems involving work and energy, always account for all forces doing work on the object, including non-conservative forces like friction.