Question

A 6.7 kg block is released from rest on a frictionless inclined plane making an angle of 25.7° with the horizontal. Calculate the time it will take this block when released from rest to travel a distance of 0.5 m along the ramp.

Answer 1

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 $\Theta =25.7^{\circ}$

Distance traveled s = 0.5 m

Initial velocity u = 0 m/sec

Acceleration of the block $a=gsin\Theta =9.8\times sin25.7^{\circ}=4.243m/sec^2$

From second equation of motion we know that $S=ut+\frac{1}{2}at^2$

So $0.5=0\times t+\frac{1}{2}\times 4.243\times t^2$

$t^2=0.2356$

t = 0.4853 sec

So time taken by block to travel 0.5 m from rest will be 0.4853 sec