A man of mass 60 kg sitting on ice pushes a block of mass of 12kg on ice horizontally with a speed of 5 `ms^(-1)`. The coefficient of friction between the man and ice and between block and ice is 0.2. If g =10 `ms^(-2)`, the distances between man and the block, when they come to rest is

Options

A. 6 m
B. 6.5 m
C. 3 m
D. 7 m

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a man of mass 60 kg sitting on ice, and he pushes a block of mass 12 kg with an initial speed of 5 m/s. We need to find the distance between the man and the block when they both come to rest due to friction.

### Step 2: Calculate the initial momentum
Using the conservation of linear momentum, we can set up the equation:
\[ m_1 v_1 + m_2 v_2 = 0 \]
where:
- \( m_1 = 60 \, \text{kg} \) (mass of the man)
- \( v_1 \) = final velocity of the man (to be determined)
- \( m_2 = 12 \, \text{kg} \) (mass of the block)
- \( v_2 = 5 \, \text{m/s} \) (initial speed of the block)

Setting up the equation:
\[ 60 v_1 + 12 \times (-5) = 0 \]
This simplifies to:
\[ 60 v_1 - 60 = 0 \]
Thus,
\[ v_1 = 1 \, \text{m/s} \]

### Step 3: Calculate the distance traveled by the man (d1)
The man will come to rest due to friction. The frictional force can be calculated using:
\[ f = \mu m g \]
where:
- \( \mu = 0.2 \)
- \( g = 10 \, \text{m/s}^2 \)

So, the frictional force on the man is:
\[ f_1 = 0.2 \times 60 \times 10 = 120 \, \text{N} \]

The deceleration \( a_1 \) caused by this friction is:
\[ a_1 = \frac{f_1}{m_1} = \frac{120}{60} = 2 \, \text{m/s}^2 \]

Using the equation of motion:
\[ v^2 = u^2 + 2as \]
where \( v = 0 \) (final velocity), \( u = 1 \, \text{m/s} \) (initial velocity), and \( a = -2 \, \text{m/s}^2 \) (deceleration):
\[ 0 = (1)^2 + 2 \times (-2) \times d_1 \]
This simplifies to:
\[ 1 = 4d_1 \implies d_1 = \frac{1}{4} \, \text{m} \]

### Step 4: Calculate the distance traveled by the block (d2)
Similarly, for the block:
The frictional force on the block is:
\[ f_2 = 0.2 \times 12 \times 10 = 24 \, \text{N} \]

The deceleration \( a_2 \) for the block is:
\[ a_2 = \frac{f_2}{m_2} = \frac{24}{12} = 2 \, \text{m/s}^2 \]

Using the same equation of motion:
\[ 0 = (5)^2 + 2 \times (-2) \times d_2 \]
This simplifies to:
\[ 0 = 25 - 4d_2 \implies 4d_2 = 25 \implies d_2 = \frac{25}{4} \, \text{m} \]

### Step 5: Calculate the total distance between the man and the block
The total distance \( D \) between the man and the block when they come to rest is:
\[ D = d_1 + d_2 = \frac{1}{4} + \frac{25}{4} = \frac{26}{4} = 6.5 \, \text{m} \]

### Final Answer:
The distance between the man and the block when they come to rest is **6.5 meters**.

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