Question

A solid sphere of mass 2 kg rolls down a fixed inclined plane of height 7 m. The rotational kinetic energy of sphere at bottom of plane is (g = 10 m/s2)​

Answer 1

Answer:

140j

Explanation:

given,

mass= 2kg

height= 7m

We know that k.e is given by 1/2 m.v²

now putting respective values,

We get,

1/2. 2. v² .........(1)

We also known that v² =u²+ 2gh ,from 2nd law

now putting value of v² in eq.1

1/2 .2. u² +2(10)(7)

1/2 .2. 0 + 14(10)..

..(;u=0)

k.e =140joules..

Answer 2

Question:-

A solid sphere of mass 2 kg rolls down a fixed inclined plane of height 7 m. Find the rotational kinetic energy of sphere at bottom of plane. (Take $\displaystyle\sf {g=10\ m\ s^{-2}.}$)

Answer:-

$\displaystyle\Large\boxed {\sf {K_r=40\ J}}$

Solution:-

Given,

• $\displaystyle\sf {M=2\ kg}$

• $\displaystyle\sf {h=7\ m}$

The rotational kinetic energy,

$\displaystyle\longrightarrow\sf{K_r=\dfrac {1}{2}I\omega^2\quad\quad\dots(1)}$

For a solid sphere, $\displaystyle\sf {I=\dfrac{2}{5}Mr^2}$

($\displaystyle\sf {r=}$ radius of sphere.)

And angular speed, $\displaystyle\sf {\omega=\dfrac {v}{r}.}$

Then (1) becomes,

$\displaystyle\longrightarrow\sf{K_r=\dfrac {1}{2}\cdot\dfrac {2}{5}Mr^2\cdot\dfrac {v^2}{r^2}}$

$\displaystyle\longrightarrow\sf{K_r=\dfrac {1}{5}Mv^2\quad\quad\dots (2)}$

By energy conservation, the total kinetic energy at the ground is obtained by the total conversion of its potential energy at the height $\displaystyle\sf {h=7\ m,}$ i.e.,

$\displaystyle\longrightarrow\sf{K_{net}=Mgh}$

Since net kinetic energy is the sum of its translational and rotational kinetic energy,

$\displaystyle\longrightarrow\sf{K_t+K_r=Mgh}$

From (2),

$\displaystyle\longrightarrow\sf{\dfrac {1}{2}Mv^2+\dfrac {1}{5}Mv^2=Mgh}$

$\displaystyle\longrightarrow\sf{\dfrac {7}{10}Mv^2=Mgh}$

$\displaystyle\longrightarrow\sf{Mv^2=\dfrac {10Mgh}{7}}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{5}Mv^2=\dfrac {2Mgh}{7}}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{5}Mv^2=\dfrac {2\times2\times10\times7}{7}}$

$\displaystyle\longrightarrow\sf{\dfrac {1}{5}Mv^2=40\ J}$

That is, from (2),

$\displaystyle\longrightarrow\Large\boxed {\sf {K_r=40\ J}}$