Newton’s Second Law Unit 3 – Lecture 3. NEWTON’S SECOND LAW STATES: ΣF = ma ΣF = net force m = mass a = acceleration.

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Presentation transcript:

Newton’s Second Law Unit 3 – Lecture 3

NEWTON’S SECOND LAW STATES: ΣF = ma ΣF = net force m = mass a = acceleration

1 st / 2 nd law summary 1 st Law Situation2 nd Law Situation Constant velocity OR at rest Speeding up, slowing down, turning Balanced forcesUnbalanced forces ΣF = 0ΣF is + ΣF is - a = 0a is + a is -

A Second Law Situation ΣF = –25 N + 15 N = –10 N This object is accelerating in the negative direction. 25 N15 N m= 15 kg –+

Find the net force. 75 kg

Find the acceleration. 75 kg

Find how far the box went after 2 seconds if it started from rest. 75 kg

Problem Solving: Finding an UNKNOWN FORCE 1.Plug mass and acceleration into ΣF = ma to find the Net Force. 2.Decompose your net force into x- and y- components. 3.Write an equation for ΣF x and ΣF y. Your equations will include the unknown force. 4. Solve for each component of the unknown force.

75 kg Find F B if the net force magnitude is 400 N.