a small block of mass m start from and slide along a frictionlessloop the loop .what should be initial height ,so the mass ushes against the to of the track with the equal force to its weight

To determine the initial height from which a small block of mass \( m \) must slide down a frictionless loop-the-loop so that it pushes against the top of the track with a force equal to its weight, we need to analyze the forces acting on the block at the top of the loop. This involves concepts from energy conservation and circular motion.

Understanding Forces at the Top of the Loop

When the block reaches the top of the loop, two main forces act on it: the gravitational force pulling it down and the normal force exerted by the track pushing it up. For the block to just maintain contact with the track at the top, the normal force must be zero, and the gravitational force must provide the necessary centripetal force to keep the block moving in a circular path.

Applying Newton's Second Law

At the top of the loop, the gravitational force \( mg \) must equal the centripetal force required to keep the block in circular motion. The centripetal force \( F_c \) can be expressed as:

• \( F_c = \frac{mv^2}{r} \)

Here, \( v \) is the velocity of the block at the top of the loop, and \( r \) is the radius of the loop. For the block to just stay in contact with the track, we set the gravitational force equal to the centripetal force:

• \( mg = \frac{mv^2}{r} \)

We can simplify this equation by canceling \( m \) (assuming \( m \neq 0 \)):

• \( g = \frac{v^2}{r} \)

Relating Velocity to Initial Height

Next, we need to find the relationship between the initial height \( h \) from which the block slides and the velocity \( v \) at the top of the loop. We can use the principle of conservation of mechanical energy, which states that the total mechanical energy (potential + kinetic) remains constant in the absence of friction.

At the initial height \( h \), the block has potential energy given by:

• \( PE_{initial} = mgh \)

At the top of the loop, the block has both potential energy and kinetic energy. The potential energy at the top of the loop (height \( 2r \)) is:

• \( PE_{top} = mg(2r) \)

The kinetic energy at the top of the loop is given by:

• \( KE_{top} = \frac{1}{2} mv^2 \)

Applying conservation of energy, we have:

• \( mgh = mg(2r) + \frac{1}{2} mv^2 \)

Substituting \( v^2 \) from our earlier equation \( v^2 = gr \) into the energy equation gives:

• \( mgh = mg(2r) + \frac{1}{2} m(gr) \)

Now, simplifying this equation:

• \( h = 2r + \frac{1}{2} r \)

This leads us to:

• \( h = \frac{5}{2} r \)

Final Result

Thus, the initial height \( h \) from which the block must start sliding down the frictionless loop-the-loop in order to exert a force equal to its weight at the top of the loop is:

• \( h = \frac{5}{2} r \)

This means that the block needs to start from a height that is two and a half times the radius of the loop to ensure it has enough energy to maintain contact with the track at the top. This analysis beautifully illustrates the interplay between gravitational potential energy and kinetic energy in a dynamic system.