Two forces at 45 pull an object distance of 14. What is work by both forces?

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To calculate the work done by two forces of 7 N each pulling at a 45-degree angle over a distance of 14 m, the forces must be resolved into components. The correct approach involves determining the resultant force using the Pythagorean theorem, yielding a force magnitude of approximately 9.9 N. The work done is then calculated using the formula W = F * d, where F is the effective force in the direction of motion. The confusion arises from miscalculating the resultant force and not applying the correct distance in the work formula. Ultimately, the accurate work done by the two forces is approximately 138.6 J.
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What is the work done by the two forces together in moving an object a distance of d = 14 m as shown in the diagram? The magnitude of each force is 7 N.
 

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evan b said:
What is the work done by the two forces together in moving an object a distance of d = 14 m as shown in the diagram? The magnitude of each force is 7 N.

Separate the forces into components. What you are interested in are the force components in the direction that the block is actually moved.

W = f * d.
 
ya, so i found force by sqrt of 7^2 + 7^2. then i multiplied it by 7 and go 69.30 J. What am i doing wrong?
 
evan b said:
ya, so i found force by sqrt of 7^2 + 7^2. then i multiplied it by 7 and go 69.30 J. What am i doing wrong?

Isn't the distance it's moved 14?
(The sqrt of 98 looks ok for the force in the direction of motion.)
 
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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