Correct Answer - Option 4 : 77 m
The correct answer is option 4) i.e. 77 m
CONCEPT:
-
Work-energy theorem: The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.
Work done, \(W = Δ KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2\)
Where m is the mass of the object, v is the final velocity of the object and u is the initial velocity of the object.
- Work-energy theorem for a variable force:
Kinetic energy, \(K =\frac{1}{2} mv^2\)
\(⇒ \frac{dK}{dt} = \frac{d(1/2mv^2)}{dt}\)
\(⇒ \frac{dK}{dt} = m\frac{dv}{dt}v = mav = Fv\)
\(⇒ \frac{dK}{dt} = F\frac{dx}{dt}\)
\(⇒ {dK} = F{dx}\)
On integrating,
\(⇒\int_{K_i} ^{K_f} {dK} = \int_{x_i} ^{x_f} F{dx}\)
\(⇒ Δ K = \int_{x_i} ^{x_f} F{dx}\)
CALCULATION:
Given that:
Mass, m = 1 tonne = 1000 kg
Initial velocity, u = 30 m/s
Final velocity, v = 0 m/s
- Here, friction does the work, and therefore work done is negative.
So, we can consider F = -k√x
From the work-energy theorem,
\(⇒ Δ K = \int_{x_i} ^{x_f} F{dx}\)
\(⇒ \frac{1}{2}m(v^2 - u^2) = \int_{0} ^{d}- k\sqrt x dx\)
\(⇒ \frac{1}{2}\times 1000 \times (0^2 - 30^2) = \int_{0} ^{d}- 1000\sqrt x dx\)
\(⇒- 450 = [- \frac{2x^{3/2}}{3} ]_{0} ^{d}\)
\(⇒450 = \frac{2d^{3/2}}{3} \)
⇒ d = 77 m
