Question

A block of mass 2Kg is at rest on horizontal table. The coefficient of friction between the block and the table is 0.1. A horizontal force of 3N is applied on the block. What is the speed of the block in m/s after it has moved a distance of 10m? ​

Answer 1

Answer:

3.33 m/s

Explanation:

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Answer 2

Answer:

The speed of the block in m/s after it has moved a distance of 10m is $\sqrt{10}$.

Explanation:

The values given in the question are,

mass of the block(m)=2kg

The initial velocity of the block(u)=0

coefficient of friction(μ)=0.1

The horizontal force applied(F)=3N

Distance moved by the block=10m

The frictional force acting of the block is,

$f=\mu N$       (1)

Where,

f=frictional force

N=weight of the block=mass×acceleration due to gravity

So,

$f=0.1\times 2\times 10=2N$     (2)

The net force acting on the block is,

$F'=F-f=3N-2N=1N$ (3)

Now we need to find the net acceleration of the block. For this, we will use equation (3) i.e.

$1N=2\times a$

$a=\frac{1}{2} ms^-^2$       (4)

From the third equation of motion we have,

$v^2-u^2=2as$         (5)

By substituting the required values in equation (5) we get;

$v^2-0^2=2\times \frac{1}{2} \times 10$

$v=\sqrt{10} ms^-^1$

Hence, the speed of the block in m/s after it has moved a distance of 10m is $\sqrt{10}$.