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 $g =10 ms ^{-2}$ )
\(a=-\mu g\)
\(\because v=u+at\)
\(0=20+(\mu\times10)\times5\)
\(50\mu=20\)
\(\mu=\frac{2}{5}\)
=0.4
Therefore , the value of coefficient of friction is 0.4
So , the correct option is (B) : 0.4
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:
\[ \mu = \frac{40}{10 \times 10} = 0.4 \]
A black body is at a temperature of 2880 K. The energy of radiation emitted by this body with wavelength between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. The Wien's constant, b = 2.88×106 nm-K. Then,

What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)
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