Question

chapter 3b friction and dynamics of uniform circular motion..

Answer 1

13,5,1/3
By energy conservation laws
1/2mv^2 =w +mgh
1/2mv^2 =fr*d +mgh
Here h = height of the block =5
And d = length of the inclination surface
Fr =friction force = meu *mg cos =1/3*mg* 12/13
Work done by the friction force = 1/3*m*10*12/13*13=40m
Now 1/2mv^2 =40m +m*10*5=90m
V^2 =underroot 180
V =36 root5



I hope it's help