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A 0.60 kg object is at rest. A 3.24 N force to the right acts on the object for 1.31 s...

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happyface22 | Student, Grade 11 | (Level 2) Honors

Posted January 15, 2012 at 8:34 AM via web

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A 0.60 kg object is at rest. A 3.24 N force to the right acts on the object for 1.31 s followed by a constant force of 4.35 N to the left is applied for 2.69 s. What is the velocity at the end of the 2.69 s

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 17, 2012 at 10:19 AM (Answer #2)

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The object has a mass of 0.6 kg. For 1.31 s a force equal to 3.24 N acts on the object towards the right followed by a force of 4.35 N towards the left for 2.69 s.

The acceleration of an object with mass m due to a force F is given by a = F/m.

For the first 1.31 s, the object is accelerated to the right at 3.24/0.6 = 5.4 m/s^2. The final velocity of the object is 0 + 5.4*1.31 = 7.074 m/s towards the right. The force of 4.35 N that is applied later, accelerates the object at 4.35/.6 = 7.25 m/s^2 towards the left. The final velocity of the object after 2.69 s is 7.074 - 7.25*2.69 = 12.4285 m/s towards the left.

After the application of both the forces the velocity of the object is 12.4285 m/s towards the left.

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