Your equation (W=Fd) is right for a conservative force (constant force), however, I would think that if there is acceleration in the +y direction, then force has to change, and therefore is not constant. Therefore you have to calculate work done by a variable force.
For example. If we toss a ball straight up into the air, the moment it leaves our hand, to the moment it hits the ground, the only acceleration it experiences is the acceleration due to gravity (-9.81 m/s^2). Its velocity changes, gradually slowing down as it reaches the highest point, then speeding up before hitting the ground, but its acceleration is constant (-9.81 m/s^2). Are you sure the problem states that the crane is lifting the rod at a constant acceleration of 1.8 m/s^2 and not a constant velocity(or speed) of 1.8 m/s^2?
The following assumes what you have stated, that acceleration is 1.8 m/s^2
Here is what I would do, although I'm not sure if it is fully correct.
Use the work-kinetic energy theorem to find work done when the force is variable. (Note: this theorem also applies for a conservative force).
Work-kinetic energy theorem:
ΔK=W
K(final) - K(initial)=W; where kinetic energy, K, is defined as 1/2(m)(v^2).
1/2(m)(v^2)(final) - 1/2(m)(v^2)(initial)=W
Getting velocity from acceleration
Take the integral, with respect to y, of your acceleration, to get velocity. (recall that the first derivative of velocity is acceleration. In other words, the antiderivative (integral) of acceleration is velocity).
Therefore, in your problem, velocity, v, equals 1.8y evaluated from y2(85meters) to y1 (0meters)
Therefore the definite integral of 1.8dy, on the interval (0, 85) is:
velocity = 1.8(85) - 1.8(0)
Velocity = 153 m/s (Note: this seems extremely fast; 153 m/s equals 342 miles per hour. 85 meters is 278 feet. This means that if an object vertically accelerates constantly (starting from ground zero) at a rate of 5.9 ft/s^2 (AKA:1.8 m/s^2), then, ignoring the downward acceleration due to gravity, the object will be traveling at a velocity of 342 miles per hour at a height of 278 feet (85 meters).)
Now plug this value into the work-kinetic energy theorem to find work done by the crane.
Work done by the crane:
1/2(m)(v^2)(final at 85 meters above the ground) - 1/2(m)(v^2)(initial at 0 meters when the object is at rest)=W
1/2(425 kg)(153 m/s)^2 - 1/2(425 kg)(0 m/s)^2=W
4974412.5 Jules= W (this amount seems unreasonably high, but when you compare it to the work done by gravity--shown below--it doesn't seem too unreasonable)
Work done by the gravitational force during the object's upward motion:
W= -Fd
W= (m)(-a)(h) ; here, height, "h", is your displacement, "d". And F=ma
W= (425 kg)(-9.81 m/s^2)(85 meters)
W= -354386.26 J
Putting it all together:
The work done by the crane (4974412.5 J) plus the work done by gravity (-354386.26 J) equals the net work.
Work (net)= 4974412.5 - 354386.26
work (net)= 4620026.25 J
Work (net)= 4.6X10^6 J (rounded to two significant figures)
