Question

A 5.0kg ball on the end of a string is whirled at a constant speed of 1.0m/s in a horizontal circle of radius 5m .what is the work done by the centripetal force during one revolution?

Answer 1

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Mass of the body(M) = 5 Kg.
• Velocity (v) = 1.0 m/s.
• Radius of circle (r) = 5m.

$\huge{\bold{\underline{Explanation:-}}}$

Finding the Centripetal force.

$\large{\bold{ F = \dfrac{Mv^2}{r}}}$

Substituting the values,

$\large{ \tt{F = \dfrac{5 \times (1)^2}{5}}}$

$\large{ \tt{F = \dfrac{ \cancel{5} \times (1)^2}{ \cancel{5}}}}$

$\large{\boxed{ \tt F = 1 \: N}}$

Now,

After one revolution the Displacement will be zero.

and the angle made by the force and displacement is 90°.

As the Centripetal force is acting inwards and displacement perpendicular to it.

Now,

Work done formula,

$\large{\boxed{ \tt{W = FS \cos \theta}}}$

Substituting the values ,

$\large{\tt{W = 1 \times 0 \times 0}}$

As cos90° = 0.

$\huge{\boxed{\boxed{ \tt{W = 0 \: J}}}}$

So, the work done by the Centripetal force is "Zero".

Answer 2

$\huge \red { \boxed{ \boxed{ \mathsf{ \mid \ulcorner Answer : \urcorner \mid }}}}$

We are given that

Mass (m) = 5.0 kg

Velocity (v) = 1.0 m/s

Radius of Circle (r) = 5 m

Displacement (s) = 0

==============================

Formula for force in circular motion is :-

$\LARGE{\boxed{\boxed{\sf{F \: = \: {\frac{mv^2}{r}}}}}}$

____________[Put Values]

F = (5) × (1)²/ 5

F = 5 × 1 / 5

F = 5/5

$\huge{\sf{F \: = \: 1 \: N}}$

$\rule{200}{2}$

Now ,

A.T.Q,

Work done is :-

$\huge{\boxed{\boxed{\sf{W \: = \: FS \: Cos \theta}}}}$

_________________[Put Values]

W = (1) × (0) × 0 [Cos90° = 0]

W = 0

$\huge{\boxed{\boxed{\sf{W \: = \: 0 \: J}}}}$