Question

A 2.0-kg ball is moving with a constant speed of 5.0 m/s in a horizontal circle whose diameter is 1.0 m. What is the magnitude of the net force on the ball

Answer 1

Answer:

100.0N

Explanation:

As the ball moves round the circle, a centripetal acceleration which is directed towards the center of the circle will keep the ball from falling off. This acceleration produces a force called centripetal force (F).

Since this is the only force acting on the ball, then the net force acting on the ball to keep it moving round the circle is the centripetal force.

F = m a         [according to Newton's second law of motion]    --------------(i)

Where;

m = mass of the ball.

a = centripetal acceleration. = $\frac{v^2}{r}$

v = speed of the ball.

r = radius of the circle.

Substitute a = $\frac{v^2}{r}$ into equation (i) as follows;

F = m x $\frac{v^2}{r}$      --------------------(ii)

From the question;

m = 2.0kg

v = 5.0m/s

r = diameter / 2     [diameter = 1.0m]

r = 1.0 / 2

r = 0.5m

Substitute these values into equation (ii) as follows;

F = 2.0 x $\frac{5.0^{2} }{0.5}$

F = 2.0 x $\frac{25.0}{0.5}$

F = 2.0 x 50.0

F = 100.0N

Therefore, the magnitude of the net force on the ball is 100.0N

Answer 2

Answer:

F = 100.0 N

Explanation:

• in order to the ball keeps moving at a constant speed in a circle, instead of moving along a straight line, there must be an acceleration that accounts for the change in direction of the ball.
• This acceleration, called centripetal, is directed at any time, towards the center of the circle.
• As any acceleration, as dictated by Newton's 2nd law, it must be produced by a force, called centripetal force.
• The magnitude of this force is related with the mass, the speed and the radius of the circle, as follows:

       $F_{c} = m *\frac{v^{2} }{r} = 2.0 kg *\frac{(5.0m/s)^{2} }{0.5m} =100.0 N$

• The magnitude of the net force on the ball is  100.0 N