A body of weight 64 N is pushed with just enough force to start it moving across a horizontal floor and the same force continues to act afterwards. If the coefficients of static and dynamic friction are 0.6 and 0.4 respectively, the acceleration of the body will be (Acceleration due to gravity = g)

The correct answer is 0.2 g.

Let's carefully analyze the problem step by step.

Given Information:
Weight of body, W = 64 N
Coefficient of static friction, μ_s = 0.6
Coefficient of dynamic (kinetic) friction, μ_k = 0.4
The applied force is just enough to start motion, and it continues to act afterwards.

Step 1: Find the mass of the body.

We know that W = mg, so:

m=\frac{W}{g}=\frac{64}{g}

Step 2: Find the applied force.

The body is pushed with "just enough force to start it moving." This means the applied force equals the maximum static friction force:

F_{app}=f_{s,max}={\mu }_s\times N

Since the floor is horizontal, the Normal force N = W = 64 N. Therefore:

F_{app}=0.6\times 64=38.4\text{N}

Step 3: Find the kinetic friction force once the body is moving.

Once the body starts moving, kinetic friction acts instead of static friction:

f_k={\mu }_k\times N=0.4\times 64=25.6\text{N}

Step 4: Apply Newton's Second Law to find the acceleration.

The same applied force (38.4 N) continues to act, but now the opposing kinetic friction is only 25.6 N. The net force on the body is:

F_{net}=F_{app}-f_k=38.4-25.6=12.8\text{N}

Now, using Newton's Second Law, F_net = ma:

a=\frac{F_{net}}{m}=\frac{12.8}{\frac{64}{g}}=\frac{12.8}{64}=0.2g

Step 5: Verify the result.

We can also express this as:

a=({\mu }_s-{\mu }_k)\times g=(0.6-0.4)\times g=0.2g

This elegant result shows that the acceleration equals the difference between the two friction coefficients multiplied by g. This is because both forces (applied force and kinetic friction) are proportional to the weight of the body (and hence to g), so the mass cancels out neatly.

Conclusion: The acceleration of the body is 0.2 g.