{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# ch06 - Chapter 6 Work and Energy 6.1 Work Done by a...

This preview shows pages 1–16. Sign up to view the full content.

Chapter 6 Work and Energy

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.1 Work Done by a Constant Force Fs W = ( 29 J joule 1 m N 1 =
6.1 Work Done by a Constant Force

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.1 Work Done by a Constant Force ( 29 s F W θ cos = 1 180 cos 0 90 cos 1 0 cos - = = =
6.1 Work Done by a Constant Force Example 1 Pulling a Suitcase-on-Wheels Find the work done if the force is 45.0-N, the angle is 50.0 degrees, and the displacement is 75.0 m. ( 29 ( 29 [ ] ( 29 J 2170 m 0 . 75 0 . 50 cos N 0 . 45 cos = = = s F W θ

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.1 Work Done by a Constant Force ( 29 Fs s F W = = 0 cos ( 29 Fs s F W - = = 180 cos
6.1 Work Done by a Constant Force Example 3 Accelerating a Crate The truck is accelerating at a rate of +1.50 m/s 2 . The mass of the crate is 120-kg and it does not slip. The magnitude of the displacement is 65 m. What is the total work done on the crate by all of the forces acting on it?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.1 Work Done by a Constant Force The angle between the displacement and the normal force is 90 degrees. The angle between the displacement and the weight is also 90 degrees. ( 29 0 90 cos = = s F W
6.1 Work Done by a Constant Force The angle between the displacement and the friction force is 0 degrees. ( 29 [ ] ( 29 J 10 2 . 1 m 65 0 cos N 180 4 × = = W ( 29 ( 29 N 180 s m 5 . 1 kg 120 2 = = = ma f s

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.2 The Work-Energy Theorem and Kinetic Energy Consider a constant net external force acting on an object. The object is displaced a distance s , in the same direction as the net force. The work is simply s F ( 29 ( 29 s ma s F W = =
6.2 The Work-Energy Theorem and Kinetic Energy ( 29 ( 29 2 2 1 2 2 1 2 2 2 1 o f o f mv mv v v m as m W - = - = = ( 29 ax v v o f 2 2 2 + = ( 29 ( 29 2 2 2 1 o f v v ax - = DEFINITION OF KINETIC ENERGY The kinetic energy KE of and object with mass m and speed v is given by 2 2 1 KE mv =

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.2 The Work-Energy Theorem and Kinetic Energy THE WORK-ENERGY THEOREM When a net external force does work on and object, the kinetic energy of the object changes according to 2 2 1 2 f 2 1 o f KE KE o mv mv W - = - =
6.2 The Work-Energy Theorem and Kinetic Energy Example 4 Deep Space 1 The mass of the space probe is 474-kg and its initial velocity is 275 m/s. If the 56.0-mN force acts on the probe through a displacement of 2.42×10 9 m, what is its final speed?

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
6.2 The Work-Energy Theorem and Kinetic Energy 2 2 1 2 f 2 1 W o mv mv - = ( 29 [ ] s F θ cos W =
6.2

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 44

ch06 - Chapter 6 Work and Energy 6.1 Work Done by a...

This preview shows document pages 1 - 16. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online