A 8.90-kg
 block of ice, released from rest at the top of a 1.32-m
-long frictionless ramp, slides downhill, reaching a speed of 2.99 m/s
 at the bottom. What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.1  N
  parallel to the surface of the ramp?

To solve this problem, we will use principles of energy conservation and work-energy theorem.

### Step 1: Work without friction
First, let's calculate the energy and speed of the block when there is no friction (already provided):

- **Mass of the block** \(m = 8.90 \, \text{kg}\)
- **Length of the ramp** \(d = 1.32 \, \text{m}\)
- **Initial speed** \(v_i = 0 \, \text{m/s}\)
- **Final speed without friction** \(v_f = 2.99 \, \text{m/s}\)

The mechanical energy is conserved. The initial potential energy (PE) of the block is completely converted into kinetic energy (KE) at the bottom:

\[
PE = KE
\]

\[
mgh = \frac{1}{2}mv_f^2
\]

where \( h \) is the vertical height of the ramp.

We can find the height using the length of the ramp and the fact that the ramp is inclined:

\[
h = \frac{v_f^2}{2g}
\]

### Step 2: With Friction
Now, when friction is acting, the work-energy theorem states:

\[
KE_{final} = PE_{initial} - W_{friction}
\]
where:
- \(W_{friction} = f \cdot d \), and \(f = 10.1 \, \text{N}\),

A 8.90-kg block of ice, released from rest at the top of a 1.32-m long frictionless ramp, slides downhill, reaching a speed of 2.99 m/s at the bottom. What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.1 N parallel to the surface of the ramp?

Explore how to calculate the final speed of a block sliding down a frictionless ramp using energy conservation principles. This detailed solution also covers the impact of a constant friction force on the final speed, applying the work-energy theorem. Suitable for Grades 11-12.

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