An object of mass m is at rest at the top of a smooth slope of height h and length l. the coefficient of kinetic friction between the object and the surface, μk, is small enough that the object will slide down the slope if given a very small push to get it started. find an expression for the object's speed at the bottom of the slope.
express your answer in terms of the variables m, μk, l, h, and appropriate constants.

The expression is

v = √(2gh - 2µk*gLcos(arcsin(h/L))

Explanation:

The expression "very small push" means that the initial Kinetic energy KE = 0.

So we have

Net Kinetic Enrgy KE = initial PE - work done

½mv² = mgh - µk*mgLcos(arcsin(h/L))

Multiplying it by 2/m

v² = 2gh - 2µk*mgLcos(arcsin(h/L))

Now taking the square root

v = √(2gh - 2µk*gLcos(arcsin(h/L))

This is the final expression

speed at the end is given as

Explanation:

As we know that here we can find the final speed of object by using work energy theorem

here when object slides down the inclined plane there will be two forces that will work on it

1) Work done by gravity

2) Work done by friction

Now we have work done by gravity

here we know that

now work done by friction force

now here we know that friction force is given as

here we know that

now we have

now the final speed is given as

so above is the speed by which it will slide at the end of inclined plane

an energy transformation flow diagram is shown

where is it ?

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.