Question

6. Force of 4 N is applied on a body of mass 20 kg.
The work done in 3rd sec. is
(2) 2 Joule
(1) 3 Joule
(4) 1 Joule
(3) 4 Joule​

Answer 1

Answer: option a) is correct

Explanation:Here, a = f/m = 4/20

= 1/5 m/s^2

Distance travelled by body in 3rd sec

S3rd = 0+ 1/5×2 (2×3-1)= 0.5m

hence, work done,

W = f× S3rd = 4 × 0.5 = 2J

Answer 2

We know that,

F = ma

4 = 20 × a

4/20 = a

1/5 = a

Now we know that, distance in nth second;

= Snth = u + ½a(2n - 1)

= Snth = 0 + ½ × 1/5 (2 × 3 - 1)

= Snth = 1/10 × (6 -1)

= Snth = 1/10 × 5

= Snth = ½

At last, work done is;

= F × Snth

= 4 × ½

= 2 J

So, option 2) 2 Joule is correct.