Question

A toy car of mass 80g is maintained to move in a horizontal circle of radius 0.8m with a velocity v ms^-1. If the centripetal force acting on it is 10 N, then the value of v in ms^-1 is

• A. 1
• B. 5
• C. 100
• D. 20
• E. 10

Answer 1

Answer: (E)

The formula for centripetal force is given by: \[ F_c = \frac{mv^2}{r} \] Where:
\( F_c \) is the centripetal force,
\( m \) is the mass of the object,
\( v \) is the velocity of the object,
\( r \) is the radius of the circular path. Given:
\( F_c = 10 \, \text{N} \),
\( m = 80 \, \text{g} = 0.08 \, \text{kg} \),
\( r = 0.8 \, \text{m} \).
Substituting these values into the formula: \[ 10 = \frac{0.08 v^2}{0.8} \] Simplifying: \[ 10 = 0.1 v^2 \] \[ v^2 = \frac{10}{0.1} = 100 \] \[ v = \sqrt{100} = 10 \, \text{ms}^{-1} \]

The correct option is (E) : \(10\)

Answer 2

Given: 
Mass of the toy car, m = 80 g = 0.08 kg
Radius of the circular path, r = 0.8 m
Centripetal force, F = 10 N

The formula for centripetal force is:
$F = \dfrac{mv^2}{r}$

Substituting the known values:
$10 = \dfrac{0.08 \cdot v^2}{0.8}$

Multiply both sides by 0.8:
$10 \cdot 0.8 = 0.08 \cdot v^2$
$8 = 0.08 \cdot v^2$

Divide both sides by 0.08:
$\dfrac{8}{0.08} = v^2$
$100 = v^2$

Taking square root:
$v = \sqrt{100} = 10$

Answer: 10