No clue how to start attacking these at allSolutionAs we nee

No clue how to start attacking these at all.

As we need to use integral formulations, we need the following basic relations:

dx/dt = v(t)

dvt/dt = a(t)

vdv/dx = a(x)

H1.) Here while projected along the incline, the object will suffer two forces for deceleration while going up, gravity and the friction.

Hence, net deceleration = gsin + g*u*cos

That is, dv/dt = -(gsin + g*u*cos)

We integrate the both sides, to get: 0 - Vo = -(gsin + g*u*cos)t

t = Vo / (gsin + g*u*cos)

We apply, x = ut + 0.5*a*t^2

To get: X = Vo^2/ (gsin + g*u*cos) - 0.5*Vo^2 / (gsin + g*u*cos) = 0.5*Vo^2 / (gsin + g*u*cos)

Hence the total distance will be twice of above = Vo^2 / (gsin + g*u*cos)

H2.) Here, the particle is at x = Xo and under acceleration = k/m*x^3 towards the centre

That is, dv/dt = -k/m*x^3; integrating the both sides, we get:

V = -(k/m*x^3)t

Hence, dx/dt = -(k/m*x^3)t Rearranging and integrating, we get:

dx/(k/m*x^3) = -dt ; That is, 0 - [m*Xo^4/4k] = -[t^2/2]

That is T = sqrt[m*Xo^4/k]

H4.) Here we have vdv/dx = -cv^n/m

hence mvdv / cv^n = -dx

integrating from Vo to v and from Xo to X

That is, m/c [(Vo^-n+2) - (V^-n+2) ] = x - xo

Further, for dv/dt = -cv^n/m

That is, mdv / -cv^n = dt; Integrating both sides, we get:

(V^-n+1) - (Vo^-n+1) = (n-1) * ct/m

Hence, V = [(Vo^-n+1) + (n-1) * ct/m]^(1/n-1)