Question

A block of 28 kgkg  kg mass m climbs an incline plane which is making  66 degreesdegrees degrees with the horizontal with an initial velocity 32 m/sm/s .?

a)How long time later does the block return to its initial position If the static, kinetic friction constant is 0.2 and gravity is 9.8m/s²?

Answer 1

Consider the motion of the block while going up along the inclined plane :

m = mass of the block = 2 kg

= angle of incline = 66 deg

F_n = normal force on the block = mg Cos

kinetic friction force acting on the block is given as

f = F_n = mg Cos

a = acceleration of the block

Force equation for the motion of the block while going upward is given as

mgSin + f = ma

- mgSin -   mg Cos = ma

a = - gSin -   g Cos

a = - (9.8)Sin66 - (0.2)(9.8) Cos66

a = - 9.75 m/s²

t = time of travel

v_f = final velocity at the top of incline = 0 m/s

v_o = initial velocity = 32 m/s

Using the equation

v_f = v_o + a t

0 = 32 + (- 9.75) t

t = 3.3 sec

d = distance traveled along the incline

Using the equation

v_f² = v_o² + 2 a d

(0)² = (32)² + 2 (- 9.75) d

d = 52.5 m

Consider the motion of the block while coming down :

Force equation for the motion of the block while going downward is given as

mgSin - f = ma

mgSin -   mg Cos = ma

a = gSin -   g Cos

a = (9.8)Sin66 - (0.2)(9.8) Cos66

a = 8.2 m/s²

t' = time of travel while coming down

v_o' = initial velocity of the block at the top = 0 m/s

Using the kinematics equation

d = v_o' t' + (0.5) a t'²

52.5 = (0) t' + (0.5) (8.2) t'²

t' = 3.6 sec

total time taken is given as

T = t + t' = 3.3 + 3.6 = 6.9 sec