Question

point A is at the bottom of a rough plane which is inclined at angle tita to the horizontal . A body of mass M is projected from A along the plane, the coefficient of friction between the body and the plane is ú. the body come to rest at a point B, a distance aX from A . obtain the expression for. (i) the work done against friction when the body move from A to B and back to A. (ii) the initial speed of the body.. (iii) the speed of the body on it returns to A​

Answer 1

The work done , when body moves along the plane

Explanation:

A body is projected from bottom of the inclined plane . When body is going up the plane .

The downward force = m g sinθ is developed due to its weight

As body is moving upwards , the force of friction will act downwards

The force of friction = μ R

here μ is the coefficient of friction ans R is the normal reaction

Thus force of friction f = μ mg cosθ

Let the acceleration upwards is a

The upward force required = m a

Thus m a = mg sinθ + μ mg cosθ

or acceleration a = g ( sinθ + μ cosθ )

The work done in moving upwards  W = F S

Thus W =  mg ( sinθ + μ cosθ ) S

here S is the displacement on the plane

When body moves down , the force of friction acts upwards

Thus m a = m g ( sinθ - μ cosθ )

The work done W = m g ( sinθ - μ cosθ ) S

As the body is projected with velocity u

which can be calculated by the relation v² - u² = - 2 a X

Here v = 0 at the highest point

Thus u = $\sqrt{2ax}$

here a = g ( sinθ + μ cosθ )

Similarly , when it moves down , the initial velocity u = 0

Thus v² - 0 = 2 a x

or  v = $\sqrt{2ax}$

here a = g ( sinθ - μ cosθ )