Work, energy, power tennis ball problem

AI Thread Summary
To calculate the work done on a tennis ball by the force F(x) = 2x + 5 as it moves from -2.5 m to 2.4 m, integration is required rather than simple multiplication. The distance traveled is 4.9 m, but plugging in the positions into the force function provides values that must be integrated over the specified interval. The initial attempt yielded an incorrect result of 48.02, highlighting the need for proper integration. The correct approach involves integrating the force function from x = -2.5 to x = 2.4 to find the total work done. Understanding this integration process is crucial for solving similar physics problems.
ferrariistheking
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Homework Statement


A force applied to a tennis ball is described by the function

F(x) = 2x + 5, with the force in Newtons and the position in meters.

How much work does the force do on a tennis ball as it moves from -2.5 m to a new position of 2.4 m?

Homework Equations


F(x)= 2x+5
force x distance= work

The Attempt at a Solution


I found the distance from -2.5m to 2.4m which is 4.9m.

Then I plugged -2.5m into x for the function which is 0.

I then plugged in 2.4m into the function which is 9.8.

With 9.8, I multiplied it by 4.9 m getting 48.02 which is incorrect

Can anybody find my error? Thanks!
 
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You need to integrate F(x) from x=-2.5 to x=2.4.
 
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andrewkirk said:
You need to integrate F(x) from x=-2.5 to x=2.4.
ohhhh I see. Thanks for the help once again.
 
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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