Question

Consider a 13 kg block which rests on the smooth surface and is subjected to a horizontal force of F=8 N. The block has an initial speed of 5 m/s and travels 24m, both directed to the right and measured from the fixed frame A. Observer B is attached to the x′ axis and is moving at a constant velocity of 55 m/s to the right relative to a stationary observer at A. Determine the work done by the force F as observed by B (in Joules).

Answer 1

Acceleration (a) = F/m = 8/13

from newton's equation of motion -

d = ut + at²/2

24 = 5t + (8/13)t²/2

t = 3.876 sec

Work done (W) = ​​​​​​

F = force

d = displacement

= angle between force and displacement

In rest frame -

displacement = 24 m

Displacement of moving frame = 55x3.876

= 213.1635 m

Diaplacement with respect to moving frame =

24 - 213.16= - 189.16 m

So work (W) = - 8x189.16

Work done (W) = - 1513.31 Joule