Question

11. A 2.5-kg block slides down a 25o inclined plane with a constant acceleration. The block starts from rest at the top. At the bottom, its velocity reaches 0.65 m/s. The length of the incline is 1.6m. a) What is the acceleration of the block?

b) What is the coefficient of friction between the plane and the block?
c) Does the result of either (a) or (b) depend on the mass of the block?

Answer 1

Step-by-step explanation:

Given

mass of Block
`m=2.5 kg`

velocity at bottom
`v=0.65 m/s`

Length of Ramp
`L=1.6 m`

inclination
`\theta =25^(\circ)`

using
`v^2-u^2=2 as`

where a=acceleration

s=displacement

v=final velocity

u=initial velocity

`0.65^2-0=2* a* 1.6`

`a=0.132 m/s^2`

(b)Block is moving downward with constant acceleration is given by

`F_(net)=mg\sin \theta -\mu mg\cos \theta`

`a=g\sin \theta -\mu g\cos \theta`

`0.132=g(\sin 25-\mu \cos 25)`

`\mu \cdot 0.906=0.4360`

`\mu =(0.4360)/(0.906)`

`\mu =0.481`

Result will remain same irrespective of mass