Asked by Anonymous
A body slides down from the rest from the top of6.4m long rough inclined plane at 30 degree with the horizontal.find the time taken by the body in reaching the bottom of the plane.
Take mu k=0.2 and g=10m/s2
Take mu k=0.2 and g=10m/s2
Answers
Answered by
Henry
Fp = Mg*sin30 = M*10*sin30 = 5M. = Force parallel to the incline in Newtons.
Fn = M*g*cos30 = M*10*Cos30 = 8.66M = Normal force = Force perpendicular to the incline.
Fk = u*Fn = 0.2 * 8.66M = 1.732M = Force of kinetic friction.
Fp-Fk = M*a.
5M-1.732M = M*a
3.268M = M*a
a = 3.268 m/8s^2
d = Vo*t + 0.5a*t^2 = 6.4.
0*t + 0.5*3.268t^2 = 6.4
1.634t^2 = 6.4
t = 1.98 s.
Fn = M*g*cos30 = M*10*Cos30 = 8.66M = Normal force = Force perpendicular to the incline.
Fk = u*Fn = 0.2 * 8.66M = 1.732M = Force of kinetic friction.
Fp-Fk = M*a.
5M-1.732M = M*a
3.268M = M*a
a = 3.268 m/8s^2
d = Vo*t + 0.5a*t^2 = 6.4.
0*t + 0.5*3.268t^2 = 6.4
1.634t^2 = 6.4
t = 1.98 s.
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