# Work done by crane

1. Apr 8, 2016

### Le_Anthony

1. The problem statement, all variables and given/known data
Had a question on a test went something like this;
A crane lifts a 425 kg bar 85m above the ground. The crane is lifting it upwards with an acceleration of 1.8 m/s^2. How much work is done by the crane?

2. Relevant equations
W=Fd

3. The attempt at a solution
Solution 1( the one i chose)
Since the crane is lifting the bar upwards with a=1.8 m/s^2, i took the total acceleration to be 9.8m/s2 - 1.8m/s2=8 m/s^2.

so 425 kg(8m/s^2)(85m)=2.9x10^5 J
(i believe this is work done by gravity.....)

Solution 2 ( what i think might be the correct answer)

Work done by CRANE is (425kg)(1.8m/s^2 )(85m)=6.5x10^4 J.
Because they asked for the work done by crane so im supposed to use 1.8m/s^2???

anyway, which solution was correct
thanks

2. Apr 8, 2016

### Staff: Mentor

Neither is correct.

Draw a Free Body Diagram for the mass being lifted. Identify the forces acting on it and write an expression for the net force. How are the net force and the acceleration related?

3. Apr 8, 2016

### drvrm

what is the logic of such equality acceleration are vectors so your 8m/s^2 is pointing down.
why not draw a free body diagram and find the force applied by crane- then calculate work

4. Apr 8, 2016

### Le_Anthony

Is it 4.2x10^5 J

5. Apr 8, 2016

### Staff: Mentor

Show your work. We won't confirm or deny guesses.

6. Apr 8, 2016

### Le_Anthony

To cancel out gravity add 9.8 to 1.8 bc it needs to move upward. 425 kg(11.6ms2)(85 m)

7. Apr 8, 2016

### Staff: Mentor

Okay, that is correct.

Suggestion: You could have gotten there quicker if you started by drawing an FBD and sorting out the forces first.

8. Apr 8, 2016

### Le_Anthony

Thanks. Kinda sad to know now haha but at least I know

9. Apr 8, 2016

### BreCheese

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)

10. Apr 8, 2016

### Staff: Mentor

This is incorrect. You must integrate acceleration with respect to time in order to obtain velocity.

The velocity is easily obtained using one of the SUVAT equations involving velocity, acceleration, and distance. The value so obtained is quite a bit more reasonable.

11. Apr 8, 2016

### BreCheese

Use this kinematical equation formula instead to get velocity: ((Vfinal)^2) -((Vinitial)^2)=2a(Δy)
((Vfinal)^2)=2(1.8 m/s^2)(85 meters)
Vfinal= 17.49 m/s
Using this equation, velocity comes out to be 17.49 m/s.
Now plug this final velocity into the work-kinetic energy theorem to find work done by the crane.

ΔKE=W
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)(17.49 m/s)^2 - 1/2(425 kg)(0 m/s)^2=W
W= 65025 J
W= 6.5X10^4 J (like what you found it to be in your original post) lol.

But now I just read through yals most recent posts and you're saying the answer is 419050 J.
Admin, which is correct? What am I doing wrong? I understand that I integrated acceleration incorrectly (because I did not integrate with respect to time. oops.). But what am I doing wrong now? should I have chosen acceleration to be something besides 1.8 m/s^2 in the kinematic equation ((Vfinal)^2) -((Vinitial)^2)=2a(Δy)?
Shucks, sorry guys. I feel like I've gotten many insights from reading yals posts; hope I didn't confuse anyone with my incorrect posts ;/

Last edited: Apr 8, 2016
12. Apr 8, 2016

### BreCheese

Ah, I think my kinematic equation is wrong, or rather, can not be applied for motion with constant acceleration in the y-direction.

13. Apr 8, 2016

### Staff: Mentor

You found the change in energy due to the change in KE, but you haven't added in change due to gravitational PE.

The crane lifts the mass through 85 meters against gravity, and it also accelerates it to some final velocity. So a change in PE and a change in KE. The total work done by the crane is the sum of both. 419 kJ is correct.