Does Calculating Block Speed Involve Integration?

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To determine the speed of a 4.1 kg block at 2.8 m, integration of the force equation F_x = ax^2 + b is necessary. The force varies with position, and the area under the curve represents the work done on the block. This work can be equated to the change in kinetic energy to solve for the final velocity. The integration process is essential for calculating the change in velocity as the block moves from 1.5 m to 2.8 m. Ultimately, the integration of the force function is crucial for finding the block's speed at the specified position.
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A 4.1kg block moving along the x-axis is acted upon by a single horizontal force that varies with the block's position according to the equation F_x = ax^2 + b, where a = 8 N/m^2, and b = -2.8 N. At 1.5 m, the block is moving to the right with a speed of 4.3 m/s. Determine the speed of the block at 2.8 m.

do i have to just integrate that formula?
 
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All that really needs to be done is to find the area underneath the curve of that equation. You can really just use any methods that please you to do this. The result would be the work done. Then set that work = the formula for kinetic energy and solve for velocity.

BTW, you're taking a physics course with calculus, right?
 


Yes, integration is involved in this problem. In order to determine the speed of the block at 2.8 m, you will need to use the equation for velocity, which involves integrating the force equation over the distance traveled. This will allow you to find the change in velocity and ultimately determine the speed at 2.8 m.
 
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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