A child starts from the rest at the top of a slide of height 4m.
(a) What is the his speed at the bottom if the slide is friction less?
(b) If he reaches the bottom, with a speed of 6 m/s. What percentage of his total energy at the top of the slide is lost as a result of friction?

(Ans: (a) 8.8 m s⁻¹, (b) 54 %)



Given:

Height of slide = h = 4 m

Value of g = 9.8 m s⁻²
  

To Find:

(a) Speed at bottom of slide =  v = ?   

(b) Percentage (%) of Total Energy Lost =  E% = ?  

Solution:


By using Law of conservation of energy

Gain in K.E = Loss in P.E

`\frac {1}{2}` m `\v²` =  m g h

or


v = `\sqrt {2gh}`


by putting the corresponding values

v = `\sqrt {2ｘ 9.8 m s⁻² ｘ4 m}`

v = `\sqrt {78.4 m² s⁻²}`

v = 8.8 m s⁻¹ -------------Ans.

Hence speed of the child at the bottom of slide is 8.8 m s⁻¹


(b) Percentage (%) of Total Energy Lost =  E% = ?  

At the top of slide the child has total of P.E.

At top E = P.E = mgh = mｘ9.8 m s ï½˜4 m = 39.2 m J ------(1)

This is the total amount of energy that child posses and m is the mass of the child.


At the bottom the P.E of the child is totally converted as K.E. So Total energy E' of the child when moving with given speed of 6 m s⁻¹:


E' = K.E = `\frac {1}{2}` m `\v²`

E' = `\frac {1}{2}` m (6m s⁻¹)²

E' = 18 m J -------(2)
{where m is mass}

Loss of Energy = E = E - E' = 39.2 m J - 18 m J = 21.2 m J {this amount of energy is lost as friction}

% loss of Energy = E% = (energy lost / Total energy) 100 %

E% = `\frac {21.2 m J}{39.2 m J}` ｘ 100 %

E% = 54 %  --------------Ans. (2)

Hence 54% of energy is lost due to friction.



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