How Much Work Is Done by Force Acting Along the X-Axis?

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The discussion revolves around calculating the work done by a force acting along the x-axis, defined by the equation Fx=(10x-2.0x^2)N, as the object moves from x=-1 to x=+6. The initial approach involved interpreting the work as the area of a triangle, with confusion about determining the base. A participant clarifies that the correct method to find the work is to compute the integral of the force function over the specified range. This integral will provide the total work done by the force during the object's movement. Understanding the integration process is essential for accurately solving the problem.
Ink
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Hi everyone

I have a question

A force acting on an object moving along the x-axis is given by Fx=(10x-2.0x^2)N
Where x is in m. how much work is done by this force as the object moves from x=-1 to x=+6

My answer was:
when it says the object moves from -1 ro +6, that means it looks like triangle and the hight will be 6m

I used this formula to getthe area of triabgle
Wab = 1/2 (base) (high), where the high is 6, but what is the base?

Here I am lost
 
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Ink said:
Hi everyone

I have a question

A force acting on an object moving along the x-axis is given by Fx=(10x-2.0x^2)N
Where x is in m. how much work is done by this force as the object moves from x=-1 to x=+6

My answer was:
when it says the object moves from -1 ro +6, that means it looks like triangle and the hight will be 6m

I used this formula to getthe area of triabgle
Wab = 1/2 (base) (high), where the high is 6, but what is the base?

Here I am lost

What they are wanting is the integral of the function over the range -1 to +6

Fx = 2*x -10*x2
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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