Answer:
So time taken by block to travel 0.5 m from rest will be 0.4853 sec
Explanation:
We have given mass of the block m = 6.7 kg
angle of inclination [tex]\Theta =25.7^{\circ}[/tex]
Distance traveled s = 0.5 m
Initial velocity u = 0 m/sec
Acceleration of the block [tex]a=gsin\Theta =9.8\times sin25.7^{\circ}=4.243m/sec^2[/tex]
From second equation of motion we know that [tex]S=ut+\frac{1}{2}at^2[/tex]
So [tex]0.5=0\times t+\frac{1}{2}\times 4.243\times t^2[/tex]
[tex]t^2=0.2356[/tex]
t = 0.4853 sec
So time taken by block to travel 0.5 m from rest will be 0.4853 sec
