Question

A locomotive of mass m starts moving so that its velocity varies according to the law v= k√s , where k is a constant and s is the distance covered. Find the total work done by all the forces acting on the locomotive during the first t seconds after the beginning of motion.



URGENT

Answer 1

ds/dt = ksqrt(s)
s^(-1/2) ds = k dt
s^(1/2) / (1/2) = kt
2sqrt(s) = kt
s = k^2 t^2 / 4
a = d2s / dt2 = k^2 / 2
dW = F dS = ma dS
Since the force is constant,
W = ma integral(dS) = ma S = (mk^2 / 2) x (k^2 t^2 / 4)

Answer 2

v = a√x

dx/dt = a√x

dx / (a√x) = dt

1/a × dx / (√x) = dt

Integrating x with limits 0 to x and dt from 0 to t

2√x / a = t

x = (at / 2)²

Displacement = (at / 2)²

Acceleration = vdv/dx

= a√x d(a√x)/dx

= a²√x × (0.5) × 1/√x

= a²/2

W = Force × Displacement

= Mass × acceleration × Displacement

= m × (a²/2) × (at/ 2)²

= (ma²/2) × (a² t²/ 4)

= ma⁴ t² / 8