Question

a force f is 2i + 3j +4k acts on a partical which is constrained to move in the xoy plane along line x=y particle moves 5 under root 2m then work done is ​

Answer 1

Answer:

25 J.

Explanation:

Since, the distances of the x and the y are same. So, we can find the distance which is moved by the particle. 5√2 = √(x^2+y^2) now, as the x=y so the distance will be 5√2 = √2s^2 which will be s=5. So, the vector by which the particle is moved will be 5i + 5j.

Hence, the amount of work done by the particle will be F.s which is                (2i + 3j +4k).(5i + 5j). So, the work done is 10+15 = 25 Joules.

Answer 2

Answer:

25 J

Explanation:

r=5

f.ds= 25 j

substitute