Three applied forces, F1= 20.0 N, F2= 40.0 N and F3= 10.0 N act on an object wit

Question

Three applied forces, F1= 20.0 N, F2= 40.0 N and F3= 10.0 N act on an object with a mass of 2.00 kg which can move along an inclined plane. The questions refer to the instant when the object has moved 0.600 m along the surface of the inclined plane in the upward direction. Neglect friction and use g=10.0 m/s^2.

What is the amount of work done by force F1 as the object moves up the inclined plane?

What is the amount of work done by force F2 as the object moves up the inclined plane?

What is the amount of work done by the force F3 as the object moves up the inclined plane?

What is the amount of work done by the gravitational force as the object moves up the inclined plane?

Explanation / Answer

The answer is 20.78 J.
I will explain this in detail since it seems to be review for your final and you should understand this.


Formula for Work is
Work = Force DOT displacement
(dot product of force and displacement)

Force2 is in the positive x direction so the Force vector is
< 40, 0>

So what is our displacement, it's .6 m. Ok so now how to represent this as a vector? It's along an inclined plane and is not straight (i.e. parallel to x or y axis) like the Force was. There's going to be a little bit in both the x and y components so how do we find this?

Remember a vector can be represented by magnitude (distance) * unit vector (direction).
We have the magnitude already, .600 m we just need to find the unit vector.

Remember each component in the unit vector is the cosine of the angle to the respected axis,
the x-component is just the cosine of the angle given since it's the angle the incline makes with the x axis.
The y-component is cosine of the angle it makes with the y axis (90 - angle given), in other words, cos(90-angle) which in this case is sin(angle).


Vector for displacement is
.600 * <cos(30deg), sin(30deg)> = <0.5196, 0.3>

So
<40, 0> DOT <.5196, .3>
= 40* .5196 + 0 * .3
= 20.78

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