What is the final speed of the crate?

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SUMMARY

The final speed of a crate with a mass of 120 kg, initially moving at 2.5 m/s, is determined through the application of net work done by forces acting on it. An astronaut applies a force of 250 N to accelerate the crate over a distance of 7 m, followed by a force of 240 N to decelerate it over 6 m. The net work done is calculated using the equation W = F1 * S1 - F2 * S2, leading to the final kinetic energy equation 0.5 * M * V2^2 = 0.5 * M * V1^2 + (F1 * S1 - F2 * S2). This results in a final speed that can be computed definitively using the provided values.

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A crate with a mass of 120 kg glides through a space station with a speed of 2.5 m/s. An astronaut speeds it up by pushing on it from behind with a force of 250 N, continually pushing with this force through a distance of 7 m. The astronaut moves around to the front of the crate and slows the crate down by pushing backwards with a force of 240 N, backing up through a distance of 6 m. After these two maneuvers, what is the speed of the crate?



Homework Equations


W=F1*deltaR+F2*deltaR2+...


The Attempt at a Solution


i can't find out how to get the speed with that? please help??
 
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Final KE = Initial KE + Network done on the crate

Let initial velocity be V1 and final velocity be V2
Let Force exerted from back be F1 and distance S1, and F2 be the force exerted from front for distance S2.

Net work done = F1*S1 - F2*S2 (minus because one is back and other is front)
0.5 M*V2^2 = 0.5*M*V1^2 + (F1*S1- F2*S2)

M, V1, F1, F2, S1, S2 are given. Use the above equation to find the final velocity.
 
thank you!
 

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