Question

19. An object of weight 4N swings freely on the end of a string. At the
HIV swings freely on the end of a string. At the bottom of the swing, the tension in the string 18
ON. If g is the magnitude of the acceleration due to gravity, the magnitude of the centrelas
bottom of the swing is:
​

Answer 1

Given:

An object of weight 4N swings freely on the end of a string. At the HIV swings freely on the end of a string. At the bottom of the swing, the tension in the string 18 ON.  

To find:

If g is the magnitude of the acceleration due to gravity, the magnitude of the centripetal acceleration of the object at the bottom of the swing is:

Solution:

From given, we have,

An object of weight 4N swings freely on the end of a string.  

At the bottom of the swing, the tension in the string 18 N.  

4 N × 4.5 = 18 N

we use the formula,

ΣF = FT – mg = mac

At the bottom of the swing, since the tension is 4.5 times the weight of the object, so we have,  

4.5mg – mg = mac

mac = 3.5mg

Therefore, the magnitude of the centripetal acceleration of the object at the bottom of the swing is 3.5mg