Answer:
0.47 N
Explanation:
Here we have a ball in motion along a circular track.
For an object in circular motion, there is a force that "pulls" the object towards the centre of the circle, and this force is responsible for keeping the object in circular motion.
This force is called centripetal force, and its magnitude is given by:
[tex]F=m\omega^2 r[/tex]
where
m is the mass of the object
[tex]\omega[/tex] is the angular velocity
r is the radius of the circle
For the ball in this problem we have:
m = 40 g = 0.04 kg is the mass of the ball
[tex]\omega =60 rpm \cdot \frac{2\pi rad/rev}{60 s/min}=6.28 rad/s[/tex] is the angular velocity
r = 30 cm = 0.30 m is the radius of the circle
Substituting, we find the force:
[tex]F=(0.040)(6.28)^2(0.30)=0.47 N[/tex]
The magnitude of the net force that the track exerts on the ball will be:
"0.47 N".
According to the question,
Mass of object, m = 40 g or,
= 0.04 kg
Radius, r = 30 cm or,
= 0.30 m
Angular velocity, ω = 60 rpm or,
= 6.28 rad/s
We know the formula,
→ Force, F = mω²r
By substituting the values, we get
= 0.040 × (6.28)² × 0.30
= 0.040 × 39.4384 × 0.30
= 0.47 N
Thus the above approach is correct.
Find out more information about friction here:
https://brainly.com/question/23161460
