First establish the summation of the forces acting int the
ladder
Forces in the x direction Fx = 0 = force of friction (Ff) –
normal force in the wall(n2)
Forces in the y direction Fy =0 = normal force in floor (n1)
– (12*9.81) –( 60*9.81)
So n1 = 706.32 N
Since Ff = un1 = 0.28*706.32 = 197,77 N = n2
Torque balance along the bottom of the ladder = 0 = n2(4 m) –
(12*9.81*2.5 m) – (60*9.81 *x m)
X = 0.844 m
5/ 3 = h/ 0.844
H = 1.4 m can the 60 kg person climb berfore the ladder will
slip
Answer:

Explanation:
By energy conservation we know that spring energy is converted into kinetic energy of the block
so we will have

so we will have

now we will have same thing for another mass 4m which moves out with speed 5v
so we have

now from above two equations we have

so we have
