Question

A particle moves along half the circumference of a circle of 1 meter radius. Calculate the work done if the force at any point is inclined at 60 degree to the tangent at that point and has 5 newtons magnitude. *​

Answer 1

Answer:

Centripetal acceleration, a

c

=k

2

rt

2

where, a

c

=

r

v

2

⇒

r

v

2

=k

2

rt

2

⇒v=krt ........(1)

Tangential acceleration, a

t

=

dt

dv

=kr.........(2)

Tangential force acting on the particle, F=ma

t

=mkr

Power delivered, P=

F

.

v

=Fvcosθ

∴ P=Fv=(mkr)×krt (∵θ=0

o

)

⟹P=mk

2

r

2

t

Answer 2

The work done is 7.854J

Given:

A particle moves along half the circumference of a circle of 1 meter

Force = 5N

Inclination = 60°

To find:

Work done by the particle with force 5N and an inclination of 60°

Solution:

We know that formula for work done by an object is

=> W = (F x cosθ) d

where,

W = work done

F = force at any given point

θ = inclination

d = distance covered by the object

In the question,

d = π ( ∵ it covers half of the circumference, i.e., $\frac{1}{2}$(2π)) = 3.1415

F = 5

θ = 60° => cos 60 = $\frac{1}{2}$ = 0.5

∴ W = (5 x 0.5) 3.1415

=> W = (2.5) 3.1415

=> W = 7.853J

Hence work done with force 5N and an inclination of 60° is 7.853J

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