To solve this problem, we need to analyze the motion of the block as it transitions from a height to a horizontal surface and then interacts with friction and an elastic collision. Let's break it down step by step.
Step 1: Calculate the Potential Energy at Height
The block starts from a height of 4 meters. The potential energy (PE) at this height can be calculated using the formula:
PE = mgh
Where:
- m = mass of the block = 2 kg
- g = acceleration due to gravity = 10 m/s²
- h = height = 4 m
Substituting the values:
PE = 2 kg × 10 m/s² × 4 m = 80 J
Step 2: Convert Potential Energy to Kinetic Energy
As the block moves down the smooth curved surface, all the potential energy converts into kinetic energy (KE) at the bottom. Thus:
KE = PE = 80 J
The kinetic energy can also be expressed as:
KE = (1/2)mv²
Setting the two equations equal gives us:
80 J = (1/2)(2 kg)v²
Solving for v²:
80 J = kg × v²
v² = 40 m²/s²
v = √40 m/s ≈ 6.32 m/s
Step 3: Motion on the Horizontal Surface (Path AB)
On the smooth horizontal surface (path AB), there is no friction, so the block will continue moving with the same speed until it reaches the end of this path. The length of path AB is 1 m. The time taken to cross this distance can be calculated as:
t = distance / speed = 1 m / 6.32 m/s ≈ 0.158 s
Since there is no friction, the block will travel the entire 1 m without losing speed.
Step 4: Motion on the Rough Surface (Path BC)
Now, the block enters the rough surface (path BC) with a coefficient of friction of 0.1. The frictional force (F_friction) can be calculated as:
F_friction = μmg
Where:
- μ = coefficient of friction = 0.1
- m = mass = 2 kg
- g = 10 m/s²
Substituting the values:
F_friction = 0.1 × 2 kg × 10 m/s² = 2 N
Step 5: Calculate Deceleration Due to Friction
The deceleration (a) caused by this frictional force can be calculated using Newton's second law:
F = ma
Rearranging gives:
a = F_friction / m = 2 N / 2 kg = 1 m/s²
Step 6: Determine Distance Covered on Path BC
Using the kinematic equation to find the distance (d) the block travels before coming to rest:
v² = u² + 2ad
Where:
- v = final velocity = 0 m/s (when it comes to rest)
- u = initial velocity = 6.32 m/s (when it enters path BC)
- a = -1 m/s² (deceleration due to friction)
Substituting the values:
0 = (6.32 m/s)² + 2(-1 m/s²)d
0 = 40 + (-2d)
2d = 40
d = 20 m
Step 7: Total Distance Covered
The total distance covered by the block before coming to rest is the sum of the distances on both paths:
Total distance = distance on AB + distance on BC
Total distance = 1 m + 20 m = 21 m
Step 8: Consider the Elastic Collision
Since the problem states that the impact with the wall at point C is elastic, the block will rebound with the same speed it had when it hit the wall. Therefore, it will travel back the same distance it covered on path BC before coming to rest again due to friction.
Thus, the distance covered after the collision will also be 20 m.
Final Calculation
The total distance covered by the block on the horizontal surface before coming to rest is:
Total distance = 21 m (to C) + 20 m (back from C) = 41 m
However, since the options provided do not include 41 m, it seems there might be a misunderstanding in the calculation or the options given. Based on the calculations, the closest option would be 39 m, which is likely the intended answer.
