What minimal amount of work is required to move a crate (weight 1,500 N) 10m on horizontal floor?

Question

A worker can apply force up to 500 N in any direction.
Coefficient of friction is μ=1/3.

Answer 1

I will assume that the coefficient of friction is the same for static and kinetic, since no difference was specified. Also, I will assume the crate is a point mass so that any upward component of force reduces the normal force and we don't have to worry about rotational motion of the crate.

The worker can over come the static friction and move the crate as long as a force is applied upward that satisfies the following inequality:

(1500-sin(th)*F)/3 where th is the angle w/r/t horizontal of the applied force.
For a force of 500 N, the inequality is satisfied for angles just greater than 0 and less than about 36.9 degrees.

So, key is to determine the smallest force required in the horizontal direction and and multiply the distance by that component of the force, which is the work done.

There are two intersting points, one where the worker uses the smallest magnitude force, which is at 18.436 degrees, where 474.34 N total force is sufficient to move the crate. Here the x component is 450 N. The second point is where the worker applies maximum of 500 N and barely overcomes friction to move the crate. Here the force is applied at 36.9 degrees. The x component of the force here is 400 N ( rounded to significant digits). So the work done is 4000 J.

Epilogue:
Alexander, I like your questions and I always look for the subtle, oops missed that, so I tested another solution by assuming the worker applied 500N at the most efficient angle, which is 18.436 degrees. He pushed the crate until he had accelerated it sufficently that it would slide to the end of the 10 m, and then reduced the normal force by the full 500 N

In this case, using conservation of energy
(500*cos(th)-
(1500-500*sin(th))/3)*d1
=1000*d2/3

and d1+d2=10
divide by 500

(cos(th)-
(3-sin(th))/3)*d1
=2*(10-d1)/3

d1=(20/3)*/
(2/3+(cos(th)-(3-sin(th))/3))

d1=9.25 m and therefore the work is 4387 J, which is more than the 4000 from above. Intuitively I expected that since the angle of application of the force that would result in the least energy loss to friction would be the winner. For some reason I just had to see it for myself.


j

Answer 2

Work = 1/3*1500*10 = 5000 N*m

However, it doesn't seem right since if the box starts at rest then the worker would not have enough force to break the frictional resistance. If it is moving at the start point, then the worker only has enough force to keep it from stopping.

Answer 3

Well actually, the minimum amount of work done can be 0...
Jst suppose one of the workerz had this crate picked up in his hands above ground. Now the direction of the force iz upwards while the distance travelled is horizontal. According to this and the formula W= (force) into (distance) into (sine of angle b/w them), the work done will be 0...

Answer 4

minimum force = W * μ = 1500 * (1/3) = 500N, Just enough for the worker to do something.