Lee pushes horizontal with a force of 79 N on a 34Kg mass for 10 m across a floor. Calculate the amount of work lee did
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Answer:
2560m or 2.56km (rounded to 3 significant figures)
Explanation:
First, list all known and desired values/variables (initial vertical velocity is 0 as the plane is kept level and vertical acceleration is just gravity):
The horizontal displacement is going to be the distance travelled, horizontally of course, once the package is released;
First thing to understand is that the vertical and horizontal components are to be dealt with separately because they don't affect each other;
Since there is no horizontal acceleration (ignoring air resistance), we simply require a velocity and time to find the horizontal displacement, using the formula v = d/t (or speed = distance/time);
What we have is the horizontal velocity but we don't have the time taken;
One thing we know is that the time elapsed for the vertical fall of 800m and for the horizontal displacement must be the same;
What we do, therefore, is find the time taken for the vertical displacement using the formula, s = ut + ¹/₂·at², since we know the vertical velocity, height and acceleration:
800 = (0)t + ¹/₂·(9.8)t²
800 = 4.9t²
t² = 163.26...
t = 12.77...
We now have the time taken for the vertical fall and the horizontal displacement, we can use this with the horizontal velocity we know already and get the horizontal displacement:
The weight of the meterstick is:
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance
from the pivot.
The torque generated by the weight of the meterstick around the pivot is:
To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:
from which we find the value of d2:
So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.