Question

A horizontal force of 400 N is exerted on a 2.0-kg ball as it rotates (at

Answer 1

Answer:

The speed of the ball is 10 meters per second.

Explanation:

The fact that ball rotates uniformly means that speed of the ball is constant and resulting acceleration is centripetal, which means that external horizontal force exerted on the ball goes towards the center of rotation. We can estimate the speed of the ball by this equation of equilibrium:

$\Sigma F_{r} = F = m\cdot \frac{v^{2}}{R}$ (Eq. 1)

Where:

$F$ - External force exerted on the ball, measured in newtons.

$m$ - Mass of the ball, measured in kilograms.

$v$ - Linear rotation speed, measured in meters per second.

$R$ - Radius of rotation, measured in meters.

We clear the linear speed in (Eq. 1):

$v^{2} = \frac{F\cdot R}{m}$

$v = \sqrt{\frac{F\cdot R}{m} }$

If we know that $F = 400\,N$, $m = 2\,kg$ and $R = 0.50\,m$, the speed of the ball is:

$v =\sqrt{\frac{(400\,N)\cdot (0.50\,m)}{2\,kg} }$

$v = 10\,\frac{m}{s}$

The speed of the ball is 10 meters per second.

Answer 2

Answer:

the speed of the ball is 10 m/s

Explanation:

Given;

magnitude of exerted force, F = 400 N

mass of the ball, m = 2 kg

radius of the circle, r = 0.5

The speed of the ball is calculated by applying centripetal force formula;

$F = \frac{mv^2}{r} \\\\v^2 = \frac{Fr}{m}\\\\v = \sqrt{\frac{Fr}{m}}\\\\ v = \sqrt{\frac{400*0.5}{2}}\\\\v = 10 \ m/s$

Therefore, the speed of the ball is 10 m/s