The velocity of a body of mass \(20 \mathrm{~kg}\) decrease from $20 \mathrm{~ms}^{-1}\( to \)5 \mathrm{~ms}^{-1}\( in a distance of \)100 \mathrm{~m}$. Force on the body is (A) \(-27.5 \mathrm{~N}\) (B) \(-47.5 \mathrm{~N}\) (C) \(-37.5 \mathrm{~N}\) (D) \(-67.5 \mathrm{~N}\)

We can find the initial and final kinetic energy of the body using the formula \(KE = \dfrac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. The initial kinetic energy is: \(KE_i = \dfrac{1}{2}(20\,\text{kg})(20\,\text{ms}^{-1})^2 = 4000\,\text{J}\) The final kinetic energy is: \(KE_f = \dfrac{1}{2}(20\,\text{kg})(5\,\text{ms}^{-1})^2 = 250\,\text{J}\)

The work done by force can be calculated as the change in kinetic energy of the body, which is given by: \(W = KE_f - KE_i\) Substituting the values, we get: \(W = 250\,\text{J} - 4000\,\text{J} = -3750\,\text{J}\)

According to the work-energy theorem, the work done by force is equal to the product of force and displacement. Thus, we can write the equation as: \(W = F \cdot d \cdot cos(\theta)\), where \(F\) is the force, \(d\) is the distance, and \(\theta\) is the angle between the force and displacement. In this case, since the body is decelerating, the force acting on the body is in the opposite direction of the displacement. So, \(\theta=180^\circ\), and \(cos(180^\circ)=-1\) Now, we can rewrite the equation as: \(-3750\,\text{J} = F \cdot (100\,\text{m}) \cdot (-1)\) \(F = \dfrac{-3750\,\text{J}}{100\,\text{m}} = -37.5\,\text{N}\) So, the force on the body is \(-37.5\,\text{N}\), which corresponds to option (C).