The problem asks about a block sliding down a frictionless ramp and then moving on a horizontal surface with friction before coming to rest. We are tasked with finding how far the block slides on the horizontal surface before it stops.

Given Information:

• Initial height of the block: $h = 5.0 \, \text{m}$
• Coefficient of friction on horizontal surface: $\mu = 0.20$
• Acceleration due to gravity: $g = 9.8 \, \text{m/s}^2$

Step 1: Determine the speed of the block at the bottom of the ramp

Using energy conservation (since there's no friction on the ramp):

• Potential energy at the top is converted into kinetic energy at the bottom.

$mgh = \frac{1}{2}mv^2$

Solving for $v$ (velocity at the bottom of the ramp):

$v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 5.0} = \sqrt{98} \approx 9.9 \, \text{m/s}$

Step 2: Determine how far the block slides on the horizontal surface

Once the block reaches the horizontal surface, the friction force will decelerate it until it comes to rest. The frictional force is given by:

$f = \mu mg$

Using Newton’s second law, the acceleration due to friction (which is deceleration) is:

$a = \frac{f}{m} = \mu g = 0.20 \times 9.8 = 1.96 \, \text{m/s}^2$

We now use the kinematic equation to find the distance $d$ the block slides before stopping:

$v^2 = u^2 + 2ad$

where $v = 0$ (final velocity), $u = 9.9 \, \text{m/s}$ (initial velocity), and $a = -1.96 \, \text{m/s}^2$ (negative because it's deceleration). Rearranging for $d$:

$0 = (9.9)^2 + 2(-1.96)d$

$d = \frac{(9.9)^2}{2 \times 1.96} = \frac{98.01}{3.92} \approx 25 \, \text{m}$

Thus, the block slides approximately 25 meters before coming to rest, which matches option (E).

Would you like further details or have any questions?

Relative questions:

1. What would happen if the coefficient of friction was reduced to 0.10?
2. How would the distance change if the initial height was doubled?
3. What if the ramp had friction? How would that affect the final distance?
4. How would the problem change if the block was given an initial push on the ramp?
5. Can we apply this concept to curved ramps, and how would the approach change?

Tip:

Conservation of energy is a powerful tool for solving many physics problems involving motion and forces. Use it when you have no information about time or specific forces along a path.