Question

Let the force or pull acting on the body make an angle of 30 o . with the horizontal. What is the value of the force to displace the body through 2m along the surface of the floor? (Cos 60 =1/2. and Cos 30 = √3/2).

Answer 1

Answer:

$\frac{10}{ \sqrt{3} } n$

Explanation:

Work done in fruit case =F.S.cosθ

=10×2×cos60

10J

Work done in second case is same as 10J=FScosθ

⇒F×2×cos30=10

F=

3

10

N

Answer 2

In first case, we are given

F = 10 N

θ = 60°

s = 2 m

So, W = Fs(cosθ)

W = 10 N × 2 m × cos 60°

= 10 N × 2 m × 1/2 (cos 60°)

= 10 J

In second case, we are given

W = 10 J

s = 2 m

θ = 30°

We have to find Force (F ), which will be given by W = Fs(cosθ).

10 = F × 2 m × cos 30°

10 = F × 2 m × √3/2 ( cos 30°)

So, F = 10/√3 N or 10√3/3 N.