Question

An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30º with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is _______?

Answer 1

Velocity at the bottom will be 16.66 m/sec

Step-by-step explanation:

We have given mass of the object m = 5.5 kg

Angle which is made by the plane
`\Theta =30^(\circ)`

Distance traveled s = 72 m

Acceleration will be equal to
`a=g(sin\Theta -\mu cos\Theta )=9.8* (sin30^(\circ)-0.35cos30^(\circ))=1.929m/sec^2`

Initial velocity u = 0 m /sec

We have to find the velocity at the bottom of the plane, that is final velocity v

From third equation of motion we know that

`v^2=u^2+2as`

So
`v^2=0^2+2* 1.929* 72=277.77`

v = 16.66 m/sec