A stone of `1kg` is thrown with a velocity of `20 ms^(-1)` across the frozen surface of lake and comes to rest after travelling a distance of `50m`. What is the force of friction between the stone and the ice?

Explanation

To solve the problem step by step, we need to find the force of friction acting on the stone as it comes to rest after being thrown across the ice.

### Step 1: Identify the given data
- Mass of the stone (m) = 1 kg
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s (since the stone comes to rest)
- Distance traveled (s) = 50 m

### Step 2: Use the kinematic equation to find acceleration
We can use the kinematic equation:
\[ v^2 = u^2 + 2as \]
Where:
- v = final velocity
- u = initial velocity
- a = acceleration
- s = distance traveled

Substituting the known values:
\[ 0 = (20)^2 + 2a(50) \]

### Step 3: Rearrange the equation to solve for acceleration (a)
Rearranging gives:
\[ 0 = 400 + 100a \]
\[ 100a = -400 \]
\[ a = -\frac{400}{100} \]
\[ a = -4 \, \text{m/s}^2 \]

### Step 4: Calculate the force of friction
Using Newton's second law, the force (F) can be calculated as:
\[ F = ma \]
Where:
- F = force of friction
- m = mass of the stone
- a = acceleration

Substituting the values:
\[ F = 1 \, \text{kg} \times (-4 \, \text{m/s}^2) \]
\[ F = -4 \, \text{N} \]

### Step 5: Interpret the result
The negative sign indicates that the force of friction is acting in the opposite direction to the motion of the stone, which is expected as friction opposes the motion.

### Final Answer
The force of friction between the stone and the ice is **4 N** (acting in the opposite direction to the motion).

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