By definition, we have that the work done is given by:

Where,
F: force in the direction of displacement
d: displaced distance
Substituting values we have:

Then, the power is given by:

Where,
t: time
Substituting values we have:


Answer:
The work done on the box is 100 Nm, and the power is 20 Nm/s.
Answer:
Explanation:
Check attachment for solution
Answer:
= 6.55cm
Explanation:
Given that,
distance = 1.26 m
distance between two fourth-order maxima = 53.6 cm
distance between central bright fringe and fourth order maxima
y = Y / 2
= 53.6cm / 2
= 26.8 cm
=0.268 m
tan θ = y / d
= 0.268 m / 1.26 m
= 0.2127
θ = 12°
4th maxima
d sinθ = 4λ
d / λ = 4 / sinθ
d / λ = 4 / sin 12°
d / λ = 19.239
for first (minimum)
d sinθ = λ / 2
sinθ = λ / 2d
= 1 / 2(19.239)
= 1 / 38.478
= 0.02599
θ = 1.489°
tan θ = y / d
y = d tan θ
= 1.26 tan 1.489°
= 0.03275
the total width of the central bright fringe
Y = 2y
= 2(0.03275)
= 0.0655m
= 6.55cm