# Calculate Distance for Work Done by Force

• Jajo
It should be obvious what constitutes a complete solution. In any event, the OP should be able to work out what the answer is from what has been said so far.In summary, to find the distance a rock was lifted, you can use the equation W = FD, where W is the work done (in joules), F is the force applied (in newtons), and D is the distance the rock was lifted (in meters). In this problem, the force of 100 Newtons and the work of 150 joules were given, and by rearranging the equation to isolate D, we can find that the rock was lifted a distance of 1.5 meters. It is important to pay attention to the units of measurement in

#### Jajo

w = Force times Distance
w = FD

4. A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??

Please i really need to understand how to do this problem, if anyone reply's please try to give step by steps

Jajo said:
w = Force times Distance
w = FD

4. A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??

Please i really need to understand how to do this problem, if anyone reply's please try to give step by steps

Well, I'm not going to do it for you! You have an equation with three variables. You know two of them, and want to find the third. How do you think you should do this?

i tried forever but i can't find a way to do this...so i have the force (100N) and i have the Work (150Joules) so i how can i get the distance

Jajo said:
i tried forever but i can't find a way to do this...so i have the force (100N) and i have the Work (150Joules) so i how can i get the distance

You have the equation W=fd. Try rearranging this to make d the subject (i.e. to get d on its own). Can you do this?

[Hint: What does the equation look like if you divide both sides by f?]

so i would divide d to both sides so it would be w/d = f right?

Jajo said:
so i would divide d to both sides so it would be w/d = f right?
No; you want to find d.

Did you read this?
cristo said:
[Hint: What does the equation look like if you divide both sides by f?]

so f = 1.5 meters or Newtons or joules

Jajo said:
so f = 1.5 meters or Newtons or joules

No. The number is correct, but which one of those units do you mean, they are not equivalent.

Ok i thought it was d...so the equation should look like this... W/F = D...so 150/100 = f...f = 1.5 meters or Newtons or Joules?

Jajo said:
Ok i thought it was d...so the equation should look like this... W/F = D...so 150/100 = f...f = 1.5 meters or Newtons or Joules?

I made a mistake in my last post, you're correct in thinking it should be D, not f that is 1.5, but still it cannot be anyone of those units they are not equivalent. So which is it Newtons, meters or joules?

"A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??"

F=100N, right?
W=150J, right?

If W=FD, and you have F and W, but want to know D, set rearrange the equation to isolate the D.. W/F=D.. 100N/150J=1.5 meters.

Why 1.5 METERS? Well, Newtons is a metric unit, and Joules is a metric unit.. And the metric unit for distance is meter. Or.. you could expand the Newtons and joules..

Newton=kg*m/(s^2)
Joule=kg*(m^2)/(s^2)

Back to the problem.. [100*kg*m/(s^2)] / [150 kg*(m^2)/(s^2)]
That should result in 1.5, and all the units except meters cancel out, therefore, 1.5 meters.

pugfug90 said:
"A force of 100 Newtons was necessry to lift a rock. A total of 150 joules of work was done. How far was the rock lifted??"

F=100N, right?
W=150J, right?

If W=FD, and you have F and W, but want to know D, set rearrange the equation to isolate the D.. W/F=D.. 100N/150J=1.5 meters.

Why 1.5 METERS? Well, Newtons is a metric unit, and Joules is a metric unit.. And the metric unit for distance is meter. Or.. you could expand the Newtons and joules..

Newton=kg*m/(s^2)
Joule=kg*(m^2)/(s^2)

Back to the problem.. [100*kg*m/(s^2)] / [150 kg*(m^2)/(s^2)]
That should result in 1.5, and all the units except meters cancel out, therefore, 1.5 meters.

Please don't provide complete solutions to homework problems.

He and everyone else has pretty much solved it already, I'm just explaining "concepts" and "theory" so that he'll understand why the answer has to be in meters.

pugfug90 said:
He and everyone else has pretty much solved it already, I'm just explaining "concepts" and "theory" so that he'll understand why the answer has to be in meters.

The OP had the correct numerical answer, however had not answered d_leet's question re the units. Besides, check out the PF guidelines: https://www.physicsforums.com/showthread.php?t=5374
guidelines said:
Under no circumstances should complete solutions be provided to a questioner, whether or not an attempt has been made.

## What is work done by force?

Work done by force is the measure of energy transferred to an object by a force acting on that object. In other words, it is the product of the force applied to an object and the distance the object moves in the direction of the force.

## What is the formula for calculating work done by force?

The formula for calculating work done by force is W = F * d, where W is work, F is force, and d is distance.

## What are the units for work done by force?

The SI unit for work done by force is joule (J). However, other commonly used units include foot-pound (ft-lb) and kilogram-meter squared per second squared (kg*m^2/s^2).

## How do you calculate distance for work done by force when the force is not constant?

In cases where the force is not constant, the work done can be calculated by integrating the force over the distance. This involves breaking the distance into small intervals and calculating the work done for each interval using the formula W = F * d. Then, the total work done can be found by summing up the work done for each interval.

## What are some applications of calculating distance for work done by force?

Calculating distance for work done by force is used in many fields, including physics, engineering, and sports. It can be used to determine the amount of energy needed to move an object, the efficiency of machines, and the power output of athletes. It is also used in the design of structures and machines to ensure they can withstand the forces applied to them.