Answer to Question #77749 in Mechanics | Relativity for bharat

Question

a block of mass 250 gm is given an initial speed of 3m/s up a plane inclined at an angle of 30 with horizontal. the coefficient of friction is 0.28. after 1.5 sec, how far is the block from its original position?

Answer 1

The acceleration of the block when it moves up
a_1=F/m=(-mg cos⁡θ-μmg cos⁡θ)/m=-g(sin⁡θ+μ cos⁡θ )=-7.28 m/s^2
Time that block will move to the rest
t_1=v_i/(-a)=3/7.28=0.41 sec
After 0.41 sec the block will move down with acceleration
a_2=-g(sin⁡θ-μ cos⁡θ )=-2.52 m/s^2
So, total displacement
S=S_1+S_2
S_1=v_i t_1+(a_1 t_1^2)/2=3×0.41-(7.28×〖0.41〗^2)/2=0.62 m
S_2=(a_2 t_2^2)/2=(-2.52×(1.5-0.41)^2)/2=-1.50 m
S=0.62-1.50=-0.88 m