Question

A body of mass $10 kg$ is moving with an initial speed of $20 m / s$ The body stops after $5 s$ due to friction between body and the floor The value of the coefficient of friction is: (Take acceleration due to gravity $g =10 ms ^{-2}$ )

• A. $0.2$
• B. $0.4$
• C. $0.5$
• D. 0.3

Hint:

The coefficient of friction can be found by equating the work done by the frictional force to the change in kinetic energy of the body.

Answer 1

The Correct Option is B

\(a=-\mu g\)
\(\because v=u+at\)
\(0=20+(\mu\times10)\times5\)
\(50\mu=20\)
\(\mu=\frac{2}{5}\)
=0.4 
Therefore , the value of coefficient of friction is 0.4
So , the correct option is (B) : 0.4

Answer 2

The work done by the frictional force is equal to the change in kinetic energy.

The frictional force \( f = \mu \times N = \mu \times mg \), where \( \mu \) is the coefficient of friction, \( m \) is the mass, and \( g \) is the acceleration due to gravity.

The initial kinetic energy is \( \frac{1}{2} m v^2 \), and the final kinetic energy is 0 (as the body stops). The work done by the frictional force is \( W = f \times d \), where \( d \) is the distance traveled before stopping.

From the equation of motion \( v_f = v_i + a t \), with \( v_f = 0 \), \( v_i = 20 \, \text{m/s} \), and \( t = 5 \, \text{s} \), we can find the acceleration \( a = \frac{v_f - v_i}{t} = \frac{0 - 20}{5} = -4 \, \text{m/s}^2 \).

Using \( F = ma \), the frictional force is \( F = 10 \times (-4) = -40 \, \text{N} \).

Now, using \( F = \mu mg \), we get:

\[ \mu = \frac{40}{10 \times 10} = 0.4 \]