Question

a body of mass moving hundred gram moving on a test track has final kinetic energy of fifty joule after traveling a distance of ten centimeter. assuming ninety percent loss of energy due to friction initial speed of the body is ​

Answer 1

Explanation:

1/2mv2 = 50J

Ki × 10/100 = 50

to find Ki

hear, Ii = me2 m = 100/1000kg.

so,

1/2mv2 = 50J

theirfore,

Ki = 500J ( Ki×10/100=50)

1/2mv2 = 500

thus,

u = root of (2×500/100)/1000

》2×5/1000

》1/100

》10ms-1

Answer 2

$\LARGE\mathfrak{\underline{\underline{\red{Question: }}}}$

A body of a mass 2kg moving on a smooth track has kinetic energy of 100 j.after traveling a distance 1 m, 19% of initial energy is lost due to air friction .the final speed of body will be?

$\LARGE\mathfrak{\underline{\underline{\pink{Solution: }}}}$

$: \implies \sf{k_j = \frac{1}{2} mv^{2} = 80 J} \\ : \implies \sf{ \frac{80}{100} \times ki = 80J} \\ : \implies \sf{ki = : \cancel\frac{{ 80} \times 100}{80} } \\ : \implies \sf{ \frac{1}{2} {mv}^{2} = 100 J} \\ \\ \mathfrak \purple{let \: the \: initial \: velocity \: energy \: be \: \sf{ki.}} \\ : \implies\sf{ki = 100J} \\ : \implies \sf{ \frac{1}{{ \not 2}} \times { \not 2 }} \times {v}^{2} = 100 \\ : \implies \sf{ {v}^{2} =100 } \\ \\ \mathfrak \purple{ \therefore initial \: velocity \: has \: 10m/s}$