3. A box starting from rest, slides down an incline which makes an angle of Thet

Question

3. A box starting from rest, slides down an incline which makes an angle of Theta above the Horizontal. The incline has a maximum height of h above the platform on which it sits which is It self H above the ground. The box leaves the end of the ramp and platform's edge and falls to the ground,the box lands a distance R from the table. Find the coefficient of friction between the box mid incline, and the total time r from the top of the incline to the floor. Theta=30.0 degree , h=0.50m.H=2.0m,R=.85 m

Explanation / Answer

let v is the speed at the edge of the ramp

vx = R/t

v*cos(30) = 0.85/t

v*t = 0.98 ---(1)

Apply, H = vy*t + 0.5*g*t^2

2 = v*sin(30)*t + 4.9*t^2

4.9*t^2 + 0.5*v*t - 2 = 0

substitute v*t value in the above equaton.

4.9*t^2 + 0.5*0.98 - 2 = 0

4.9*t^2 = 1.51

t = 0.555 s

from equation 1

v = 0.98/0.555

= 1.765 m/s

now, apply v^2 - u^2 = 2*a*s

v^2 - 0^2 = 2*a*h*sin(30)

a = v^2/(2*h*0.5)

= 1.765^2/(2*0.5/0.5)

= 1.56 m/s^2

now apply, a = g*sin(30) - mue_k*g*cos(30)

1.56 = 9.8*0.5 - mue_k*9.8*cos(30)

mue_k = (4.9-1.56)/(9.8*cos(30))

= 0.394 <<<<<<----------Answer

time taken to slide, t = (v-u)/a

= (1.765-0)/1.56

= 1.13 s

total time = 0.555 + 1.13

= 1.686 s <<<<<--------Answer

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