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

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

Step-by-step 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* sin25.7^(\circ)=4.243m/sec^2`

From second equation of motion we know that
`S=ut+(1)/(2)at^2`

So
`0.5=0* t+(1)/(2)* 4.243* 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