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A 32 kg block slides up a hill to a height of 6.4 m. If 266 J of thermal energy are generated, how fast was the block going at the bottom of the hill

Answer :

Answer:

Speed of the block at the bottom of the block is 11.91 m/s.

Explanation:

Energy can neither be created nor be destroyed but it can change its form from one to another such as kinetic energy to potential energy or vice a versa.

Applying conservation of energy to the given problem :

Kinetic Energy = Potential Energy + Thermal Energy

[tex]\frac{1}{2}mv^{2} = mgh\ + 266[/tex]

Here m is mass of the block, h is height, g is acceleration due to gravity and v is speed of the block.

Rearrange the above equation in terms of v ;

 [tex]v = \sqrt{2(gh\ + \frac{266}{m})}[/tex]

Substitute 32 kg for m, 6.4 m for h and 9.8 m/s² for g in the above equation.

[tex]v = \sqrt{2(9.8\times6.4\ + \frac{266}{32})}[/tex]

v = 11.91 m/s

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