Question

Q:A 2kg box slides down a ramp a distance of two meters before it reaches the ground.The ramp has an angle 30 degree.The coefficient of kinetic friction for ramp is 0.1 .What is the work done by gravity on box ?

(a)35 J
(b)20 J
(c)10 J
(d)40 J​

Answer 1

Explanation:

 block is moving down the plane with constant speed

So here friction force on the block is counter balanced by weight along the incline plane while it is sliding down

                mgsinθ=Ff

 now when block is pushed up the plane with initial speed "v"

then it has two forces along the incline plane opposite to the velocity

 1. friction force

 2. component of weight along the incline

so net force along the incline opposite to velocity

Fnet=−mgsinθ−Ff=−2mgsinθ 

deceleration of the block is given by      

a=−2gsinθ

we have

vf2−vi2=2ad

0−v2=2×(−2gsinθ)×d

d=4gsinθv2  

Answer 2

Explanation:

Work is determined using the equation W=FdcosΘ. Here, F is the force applied, d is the displacement of the object, and cosΘ is the angle of the force relative to the movement of the object. Since gravity is acting on the box, we can solve for the force of gravity causing the movement of the box. Note that in this case Θ refers to the angle between gravity and the box's path; thus, the angle will be 60o, rather than 30o.

W=FdcosΘ=(2kg)(10ms2)(2meters)cos(60o)

W=20J.

Notice how the work done by gravity is equal to the potential energy of the box at the top of the ramp. This is because mechanical energy is conserved in the system; thus, we can set the two equations equal to each other.

ΔUg=FdcosΘ

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