Question : A block is gently placed at the top of an inclined plane 6.4 m long . Find the time taken by the block to slide down to the bottom of the plane . The plane makes an angle 30° with the horizontal. Coefficient of kinetic friction between block and plane is 0.2. [Take g=10 m/s²]
Doubt by Krish
Solution :
length of the inclined plane (s) =6.4 m
Angle of inclination (θ) = 30°
Coefficient of kinetic friction (µ) = 0.2
initial velocity (u) = 0 m/s
We know, acceleration of a body moving down the inclined plane is given by
a = g(sinθ-µcosθ)
a = g(sinθ-µcosθ)
a = 10(sin30°-0.2×cos30°)
a = 10([1/2]-0.2×[√3/2])
a = 10[0.5-0.2×0.866]
a = 10[0.5-0.1732]
a = 10[0.3268]
a = 3.268 m/s²
a = 10([1/2]-0.2×[√3/2])
a = 10[0.5-0.2×0.866]
a = 10[0.5-0.1732]
a = 10[0.3268]
a = 3.268 m/s²
Now, Using 2nd Equation of Motion
s=ut+½at²
s=(0)t+½at²
s=½at²
2s/a=t²
t=[2s/a]1/2
=[(2×6.4)/3.268]1/2
=[12.8/3.268]1/2
= [3.9168]1/2
= 1.979
s=(0)t+½at²
s=½at²
2s/a=t²
t=[2s/a]1/2
=[(2×6.4)/3.268]1/2
=[12.8/3.268]1/2
= [3.9168]1/2
= 1.979
= 1.98 s
Hence, the time taken by the plane to slide down to the bottom of the plane will be approximately 1.98 seconds.