Question

A block is released from rest at point p and slides along the frictionless track shown. at point q, its speed is:​

Answer 1

height travels from p to q

(h1-h2 )

(potienal energy ) p =Q ( kinetic energy)

mg( h1 - h2) = 1/2mv^2

2g(h1-h2) = v^2

###

(√2g ( h1-h2 ) = v ) ANSWER

Answer 2

Concept:

Regardless of any internal changes that may occur, including energy dissipating in one form and reappearing in another, the total energy of an isolated system remains constant.

Given:

Block is released from point p

Track is frictionless

To find:

speed at point q

Solution:

Let the mass of the object is m.

At point P,

Kinetic energy=$0$

Potential energy= $mgh_{1}$

Total energy, $T_{P}$$= mgh_{1}$

At Point Q,

Let the velocity of the object be v

Kinetic energy= $\frac{1}{2} mv^{2}$

Potential energy=$mgh_{2}$

Total energy, $T_{Q}$= $\frac{1}{2} mv^{2} + mgh_{2}$

Using the law of conservation of energy

$T_{P} =T_{Q}$

$mgh_{1}$= $\frac{1}{2} mv^{2} + mgh_{2}$

$v= \sqrt{2g(h_{1} -h_{2} } )$

The speed at point Q will be, $v= \sqrt{2g(h_{1} -h_{2} } )$

                                                                   #SPJ2