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  • A block of mass 3.00 kg moving at a speed of 8.0 m/s accelerates at 6.00  <img alt="\mathrm{~m} / \mathrm{s}^2" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B

A block of mass 3.00 kg moving at a speed of 8.0 m/s accelerates at 6.00  \mathrm{~m} / \mathrm{s}^2 for 7.00 s. Compute its final kinetic energy.

 

Option: 1

3625 Joules 


Option: 2

3750 Joules 


Option: 3

5780 Joules

 


Option: 4

3268 Joules


Answers (1)

The final velocity of the block can be calculated using the formula:

v=u+a t

where a is the acceleration of the block and t is the time for which the acceleration is applied.

In this case, the acceleration of the block is 6.00 \mathrm{~m} / \mathrm{s}^2 and the time for which it is applied is 7.00 s. Therefore, the final velocity of the block is:

v =8 \mathrm{~m} / \mathrm{s}+6.0 \mathrm{~m} / \mathrm{s}^2 \times 7.0 \mathrm{~s}

=50 \mathrm{~m} / \mathrm{s}

The final kinetic energy of the block is:

K E_{\text {final }} =\frac{1}{2} \times 3.0 \mathrm{~kg} \times(50 \mathrm{~m} / \mathrm{s})^2 \\ \\

                  =3750\: \mathrm{Joules}

Therefore, the final kinetic energy of the block is 3750 J.

Posted by

Sumit Saini

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