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A block of mass 4 kg is kept over a roug...

A block of mass 4 kg is kept over a rough horizontal surface. The coefficient of friction between the block and the surface is 0.1 At, t=0, `(3hati)(m/s)` velocity is imported to the block simultaneously `(-2hati)N` force starts acting on it. Its displacement in first 5s is

A

`8hati`

B

`-8hati`

C

`3hati`

D

`-3hati`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we will follow these steps: ### Step 1: Identify the forces acting on the block The block has a mass of \( m = 4 \, \text{kg} \). It is subjected to: - An initial velocity \( \vec{u} = 3 \hat{i} \, \text{m/s} \). - A force \( \vec{F} = -2 \hat{i} \, \text{N} \) acting in the opposite direction. - A frictional force due to the rough surface. ### Step 2: Calculate the normal force and the frictional force The normal force \( N \) acting on the block is equal to its weight since it is on a horizontal surface: \[ N = mg = 4 \, \text{kg} \times 10 \, \text{m/s}^2 = 40 \, \text{N} \] The frictional force \( f \) is given by: \[ f = \mu N = 0.1 \times 40 \, \text{N} = 4 \, \text{N} \] Since the frictional force acts in the opposite direction to the motion, it will also be negative. ### Step 3: Calculate the net force acting on the block The total force acting on the block is the applied force plus the frictional force: \[ F_{\text{net}} = -2 \, \text{N} - 4 \, \text{N} = -6 \, \text{N} \] ### Step 4: Calculate the acceleration of the block Using Newton's second law \( F = ma \): \[ -6 \, \text{N} = 4 \, \text{kg} \cdot a \] Solving for \( a \): \[ a = \frac{-6 \, \text{N}}{4 \, \text{kg}} = -\frac{3}{2} \, \text{m/s}^2 \] ### Step 5: Determine the time until the block comes to rest Using the equation of motion \( v = u + at \), where \( v = 0 \) (final velocity when the block comes to rest): \[ 0 = 3 \, \text{m/s} + \left(-\frac{3}{2} \, \text{m/s}^2\right) t \] Rearranging gives: \[ \frac{3}{2} t = 3 \quad \Rightarrow \quad t = \frac{3 \times 2}{3} = 2 \, \text{s} \] ### Step 6: Calculate the displacement during the first 2 seconds Using the equation for displacement \( s = ut + \frac{1}{2}at^2 \): \[ s = (3 \, \text{m/s})(2 \, \text{s}) + \frac{1}{2}\left(-\frac{3}{2} \, \text{m/s}^2\right)(2 \, \text{s})^2 \] Calculating each part: \[ s = 6 \, \text{m} - \frac{1}{2} \times \frac{3}{2} \times 4 = 6 \, \text{m} - 3 \, \text{m} = 3 \, \text{m} \] ### Final Answer The displacement of the block in the first 5 seconds is: \[ \boxed{3 \hat{i} \, \text{m}} \]
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