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
I will mark brain list please answer question

Answer 1

m= 100g= 0.1kg

h=10cm= 0.01m

50-1/2mv^2= 10/100×mgh

1/2mv^2= 50-0.1×0.1^10×0.01= 50-0.001= 49.999

v^2=2×49.999/0.1=999.98

v=31.62m/s

Answer 2

$\bf{m= 100g= 0.1kg}\\ \bf{</p><p>h=10cm= 0.01m} \\ \bf{50-\frac{1}{2}mv^2=\frac{10}{100}×mgh} \\ \bf{\frac{1}{2} mv^2= 50-0.1×0.1^{10}×0.01} \\ \bf{= 50-0.001= 49.999} \\ \bf{v^2=\frac{2×49.999}{0.}=999.98} \\ \bf{</p><p>v=31.62m/s}$