How much work would it take to push a # 1 kg # weight up a # 3m # plane that is at an incline of # pi / 4 #?

Question

How much work would it take to push a #   1 kg  # weight up a # 3m # plane that is at an incline of #  pi  /  4 #?

Sebastian Acosta · Accepted Answer

This requires 20.8 J of work done.

The definition of work is #W = F * Deltad costheta# but care must be taken in using the formula as the angle #theta# must be the angle between the displacement #(Deltad)# and the force against which your are doing work. Here, the force we work against is the force of gravity, so the angle in the above formula is the one between the incline and the vertical. This means it is not #pi/4#, as this is measured relative to the horizontal, but the compliment of #pi/4# which in this case also happens to be #pi/4#! But keep this in mind for next time, when the given angle is something else! So, if the mass of the object is #m#, the force of gravity on it is #mg#, which equals 9.8 N in this case. #W=F*Deltad costheta = 9.8 xx 3 xx cos(pi/4)= 20.8 J# since #cos(pi/4) = 1/sqrt2#

Amelia Adams · Answer

undefined To find the work done in pushing the weight up the incline, you can use the formula:

$$ W = F \times d \times \cos(\theta) $$

Where:
- $ W $ is the work done (in joules)
- $ F $ is the force applied parallel to the incline (in newtons)
- $ d $ is the displacement along the incline (in meters)
- $ \theta $ is the angle of the incline (in radians)

Given:
- $ m = 1 $ kg (mass of the weight)
- $ d = 3 $ m (displacement along the incline)
- $ \theta = \frac{\pi}{4} $ radians

To find $ F $, you can resolve the weight force ($ mg $) into two components: one parallel to the incline ($ mg \sin(\theta) $) and one perpendicular ($ mg \cos(\theta) $). Since the weight is being pushed up the incline, the force opposing the motion is the parallel component ($ mg \sin(\theta) $). Thus:

$$ F = mg \sin(\theta) $$

Substituting the given values:

$$ F = (1 \, \text{kg}) \times (9.81 \, \text{m/s}^2) \times \sin\left(\frac{\pi}{4}\right) $$

$$ F \approx 9.81 \, \text{N} \times 0.707 $$

$$ F \approx 6.94 \, \text{N} $$

Now, plug the values into the work formula:

$$ W = (6.94 \, \text{N}) \times (3 \, \text{m}) \times \cos\left(\frac{\pi}{4}\right) $$

$$ W \approx 6.94 \, \text{N} \times 3 \, \text{m} \times 0.707 $$

$$ W \approx 14.71 \, \text{J} $$

So, it would take approximately 14.71 joules of work to push the 1 kg weight up a 3 m plane at an incline of $ \frac{\pi}{4} $ radians.