Question

A block with mass m = 17.2 kg slides down an inclined plane of slope angle 13.8^o with a constant velocity. It is then projected up the same plane with an initial speed 4.05 m/s. How far up the incline will the block move before coming to rest?

Answer 1

in OAB ,

OA = hypoteneuse = mg = weight of the block

OB = adjacent

AB = opposite

Using trigonometric formula ::

Sin = opposite/hypoteneuse = AB/OA = AB/mg

so AB = mgSin

similarly using

Cos = Adjacent/hypoteneuse = OB/OA = OB/mg

so OB = mgCos

Perpendicular to incline , the force equation can be given as ::

F_n = mg Cos                              eq-1

lets assume the coefficient of friction as ''"

then frictional force is given as ::

f = F_n

f = mg Cos               (Using eq-1)

Parallel to incline , the force equation can be given as ::

mg Sin = f                               since block moves at constant velocity

mg Sin = mg Cos  

tan =

tan13.8 =

= 0.246

while moving up ::

net force acting on the block parallel to incline is given as ::

F_net = mg Sin + f

F_net = mg Sin + mg Cos  

ma = mg Sin + mg Cos                      since F_net = ma

a = g (Sin + Cos )

a = (9.8) (Sin13.8 + (0.246) Cos13.8 )

a = 4.68 m/s²         in opposite direction of motion of block

V_i = initial velocity of block = 4.05 m/s

V_f = final velocity = 0 m/s             since block comes to rest finally

d = distance travelled.

Using the equation ::

V_f² = V_i² + 2 a d

0² = (4.05)² + 2 (- 4.68) d              (acceleration is negative since velocity is in opposite direction of acceleration)

d = 1.75 m