Question

A body ascends a slope with a speed of 10 m/s if 10% of energy of the body its lost due to friction the height to which the body will rise is take g=10m/s

Answer 1

$\Large\boxed{\sf{\quad4.5\ m\quad}}$

The initial speed (speed at the bottom of the slope),

• $\sf{u=10\ m\ s^{-1}}$

Let the height (vertical) until which the body rises be $\sf{h.}$

Since 10% of the energy is lost by the time the body reaches the height due to friction, the total energy of the body at the height $\sf{h}$ will only be 90% of its total energy during the journey, i.e., at the bottom.

Remember, law of conservation of mechanical energy can't be applied here because frictional force acts on the body. The law can be applicable only if all the forces acting on the body are conservative.

At the bottom, the body has kinetic energy but no potential energy,

• $\sf{K=\dfrac{1}{2}\,mu^2}$

At the maximum possible height $\sf{h,}$ the body stops so it has no kinetic energy there, but potential energy,

• $\sf{U=mgh}$

As said earlier,

$\longrightarrow\sf{U=\dfrac{9}{10}\ K}$

$\longrightarrow\sf{mgh=\dfrac{9}{10}\cdot\dfrac{1}{2}mu^2}$

$\longrightarrow\sf{gh=\dfrac{9}{20}\,u^2}$

$\longrightarrow\sf{h=\dfrac{9u^2}{20g}}$

Taking $\sf{g=10\ m\ s^{-2},}$

$\longrightarrow\sf{h=\dfrac{9\times100}{20\times10}}$

$\longrightarrow\sf{\underline{\underline{h=4.5\ m}}}$