to push a 25 kg crate up a 27 incline, a worker exerts a force of 120 N, parallel to the incline. As the crate slides 3.6 m, how much work is done on the crate by a) the worker b) force of gravity c) normal force
Wc = mg = 25kg * 9.8N/kg = 245N. = Weight of crate.
Fc = 245N @ 27deg.
Fp = 245sin27 = 111.2N.=Force parallel to incline.
Fv = 245cos27 = 218.3N. = Force perpendicular to incline = Normal.
a. W = Fd = 120 * 3.6 = 432J.
b. W = Fc*d=Fc*h=245 * 3.6sin27 = 400J.
c. W=Fv * d = 218.3 * 3.6sin27 = 357J.
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a) The worker exerts a force of 120 N parallel to the incline. Considering only the component of this force that is parallel to the motion, we can determine the work done by the worker using the formula:
Work = Force × Distance × cos(theta)
where theta is the angle between the force and the direction of motion. Since the force is parallel to the incline, the angle between the force and the direction of motion is 0 degrees, so cos(0) = 1. Therefore, the work done by the worker is:
Work = 120 N × 3.6 m × cos(0) = 120 N × 3.6 m × 1 = 432 J
So, the work done by the worker is 432 joules.
b) The force of gravity is acting perpendicular to the incline, so no work is done by the force of gravity. This is because the displacement is in the direction of the incline while the force of gravity is perpendicular to it.
Therefore, no work is done by the force of gravity on the crate.
c) The normal force is acting perpendicular to the incline, so no work is done by the normal force.
Therefore, no work is done by the normal force on the crate.
I hope that answers your question! If not, I can always try to come up with a less "inclining" answer!
To determine the work done on the crate, we need to consider the forces involved and the displacement of the crate.
a) Work done by the worker:
The work done by an individual force is given by the formula W = F * d * cos(θ), where W is the work done, F is the force applied, d is the displacement, and θ is the angle between the force and the direction of displacement.
In this case, the worker exerts a force of 120 N parallel to the incline, so the angle θ between the force and the displacement is 0 degrees (since they are parallel). The displacement, d, is the distance the crate slides, which is given as 3.6 m.
So, W = F * d * cos(θ) = 120 N * 3.6 m * cos(0°) = 120 N * 3.6 m * 1 = 432 J (joules)
Therefore, the work done by the worker on the crate is 432 joules.
b) Work done by the force of gravity:
The force of gravity is acting vertically downwards. Since the displacement of the crate is along the incline, the angle θ between the force of gravity and the displacement is 90 degrees (since they are perpendicular).
In this case, the work done by the force of gravity is W = F * d * cos(θ) = m * g * d * cos(90°), where m is the mass of the crate (25 kg) and g is the acceleration due to gravity (9.8 m/s²).
Since cos(90°) = 0, the work done by the force of gravity will be 0 joules.
Therefore, the work done by the force of gravity is 0 joules.
c) Work done by the normal force:
The normal force acts perpendicular to the incline and is not doing any work because there is no displacement in the direction of the normal force. Therefore, the work done by the normal force is also 0 joules.
In summary:
a) The worker's work = 432 J
b) The force of gravity's work = 0 J
c) The normal force's work = 0 J