Review velocity is (change in position)/time elapsed acceleration is (change in velocity/time elapsed Changes can be in magnitude or in direction or both.

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Presentation transcript:

Review velocity is (change in position)/time elapsed acceleration is (change in velocity/time elapsed Changes can be in magnitude or in direction or both total force =mass x acceleration Kinds of forces: gravitational,electromagnetic Different kinds of forces can add to give total force. Near the earth’s surface gravitational force = Mass x g down (1/2)mv 2 + mgh = constant (conservation of energy)‏ kinetic gravitational potential

Review page 2: When one object exerts a force F on another object through a distance d along the direction of the force, an amount of WORK Fd is done by the first object on the second and an amount of energy Fd is transferred from the first object to the second. Newton’s third law says that when one object exerts a force F on a second object, then the second object exerts a force –F(same magnitude, opposite direction) on the first object.

Suppose the 1 kg block being pulled across the table is accelerating at 8m/s 2 and that the force pulling the block is 10 newtons. What is the magnitude and direction of the friction force on the block? A. 9.8 newtons down B. 10 newtons backward C. 2 newtons backward D. 2 newtons forward Some practise questions.

Answer C. ma=1kg x 8m/s 2 = 10 newtons -magnitude of backward friction force

If the same 1kg block (pulled with a 10 newton force) was pulled through 1/2 meter while accelerating at 8 m/s 2 how much thermal energy was generated? A. 5 joules B. 4 joules C. 1 joule D. 4.9 joules

Answer C. Friction force on the table x ½ meter =1joule Notes: Total work done by pulling the block = 10 newtons x ½ meter = 5 joules So the kinetic energy of the block after 1 s must have been 4 joules. See if you can show that the final kinetic energy of the block is 4 joules by finding the final velocity of the block and using the definition KE = ½ mv 2

If you assume that the moon moves in a circle around the earth at constant speed (which is approximately true), what is the direction of the velocity, acceleration and total force on the moon at any moment? v a F A.tangent to the circle straight toward earth straight toward earth earth B. tangent tangent toward earth C. toward earth toward earth toward earth D. tangent toward earth tangent

Answer: A.

When I drop an elastic ball and it hits the floor and bounces, what is the direction of its velocity, acceleration and the total force on it when it is at its lowest point (in contact with the floor)? v a F A. d d d B. 0 u d C. 0 u u D. u u u E. 0 d d

Answer: C.

When I dropped a rubber ball from one meter, it was observed to rise to 0.75m after the first bounce. How was energy conserved? Answer: A. Kinetic energy of ball increased. B. The thermal energy of the floor and the ball increased. C. The kinetic energy of the earth increased. D. The ball was spinning.

Answer B.

I push a cart up with a constant force up a slope to a height of 1/4 meter. The mass of the cart is 1/4 kg. The cart moves 1/2 meter along the table as I do this. After I let it go, the cart continues up the slope until it reaches a height of 1/2 meter and then starts back down. What was the magnitude of the force with which I pushed the cart? A. 2.45/√5 newtons B. 4.9/ √5 newtons C. 4.9 newtons D newtons

B. Total energy gain =(1/4kg)x(9.8m/s 2 )x(1/2 m) =Fx (distance along force) =Fx√((1/2) 2 + (1/4) 2 )=Fx(√5/4) Solve for F

In the same situation, at what speed was the car moving when it left my hand? A. √4.9 m/s B. √2.45 m/s C. √(4.9√5) m/s D. √9.8 m/s

Answer A. (¼ m)x(9.8m/s 2 )(1/4 kg)=(1/2)(1/4 kg) v 2 Solve for v.

A baseball player hits a ball. The velocity as the ball leaves the bat is 20 m/s up and 40 m/s horizontal. How much time does it take for the ball to reach its highest point? Answer: A. 40/9.8 sec B. 20√5/9.8 sec C. 20/9.8 sec D. 9.8/20 sec

Answer: C. Ball loses 20m/s in the time t while decelerating at 9.8m/s 2 so 20m/s=(9.8m/s 2 )xt Solve for t.

Same baseball problem: How far away from the batter in the horizontal direction does the ball land? A. 800/9.8m B. 1600/9.8 m C. 400/9.8 m D. 3200/9.8 m

Answer: B Time x horizontal velocity Time is twice time to get to the highest point.