Question

A 0.50 kg object moves in a horizontal circular track of radius of 2.5 m An external force of 3.0 N, always
tangent to the track, causes the object to speed up as it goes around. The work done by the external force as
the object makes one revolution is :
(A) 24 J
(B) 47 J
(C) 59 J
(D) 94 J
​

Answer 1

The  work done by the external force as  the object makes one revolution is 47 J.

Explanation:

Given that,

Mass of the object, m = 0.5 kg

Radius of the track, r = 2.5 m

Force acting on the object, F = 3 N

Let W is the work done by the external force as  the object makes one revolution. Its formula is given by :

$W=F\times d$

$W=F\times 2\pi r$

$W=3\times 2\pi \times 2.5$

W = 47.12 Joules

or

W = 47 J

So, the  work done by the external force as  the object makes one revolution is 47 J.

Learn more :

Topic : Centripetal force

https://brainly.in/question/4037970

Answer 2

Answer:

above answer is correct

Explanation:

can confirm