A 0.50 kg object is at rest. A 3.29 N force

mustang

Problem 10.
A 0.50 kg object is at rest. A 3.29 N force to the right acts on the object during a time interval of 1.60 s.
At the end of this interval, a constant force of 4.07 N to the left is applied for 2.75 s.
b. What is the velocity at the end of the 2.75 s? Answer in m/s.
Note: Is the answer -22.385 m/s since in the question it says the force to the left?

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HallsofIvy

Homework Helper
Okay, what have YOU done on this problem so far? What formulas do you think might apply?

mustang

I believe that the impulse-momentum theorem can be used here to slove the problem.

HallsofIvy

Homework Helper
F= ma. You know the force and you know the mass so you can calculate the acceleration. Of course, the change in velocity is "acceleration * time".

To answer your original question, no, the "answer" is not -22.385 m/s. -22.385 m/s is the change in speed due to the second force. What was the speed when the second force was applied?

mustang

Would you add 2.75s to 1.60s to get 4.35s? From that would you then divide 0.50kg from 4.07N, to get 8.14 which is mulitplied by 4.35 to get the answer 35.409 m/s.

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