Question #77749
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?
Expert's answer
The acceleration of the block when it moves up
a_1=F/m=(-mg cosθ-μmg cosθ)/m=-g(sinθ+μ cosθ )=-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(sinθ-μ cosθ )=-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
a_1=F/m=(-mg cosθ-μmg cosθ)/m=-g(sinθ+μ cosθ )=-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(sinθ-μ cosθ )=-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
