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# Newton's Laws of Motion: 5 Problems

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1. A 1-kg block of wood is dropped from a position 2.4 meters above a resting 0.05-kg ball. Shortly thereafter, the ball is shot horizontally at a velocity of 28 m/s. If the wood is 8.4 meters away from the ball in the horizontal direction, determine the distance (to two decimal places) the wood falls from its original position until impact? Ignore air resistance and use -9.8 m/s/s for the acceleration due to gravity.

2. A hockey puck is hit on a frozen lake and starts moving with a velocity of 11.5 m/s. After 7.45 seconds, its velocity is 5.67 m/s. What is the average value of the coefficient of friction between the puck and the ice? Enter your answer accurate to the fourth decimal place.

(Referring to the previous problem.) How far (in meters) does the puck travel during this 7.45 second time interval?

3. A 2835-Newton bobsled leaves a horizontal track with a speed of 30.5 m/s. It then moves onto a horizontal track lightly sprinkled with sawdust and comes to a stop in a distance of 144 meters. What is the coefficient of kinetic friction between the runners of the sled and the sawdust-covered surface? Enter your answer accurate to the fourth decimal place.

4. Dexter Eius is walking through the cafeteria when he slips on some spilled milk and falls to the floor. Upon hitting the floor, he skids to a stop with an acceleration of -5.17 m/s/s. Dexter weighs 663 Newtons. Determine the coefficient of friction between Dexter and the floor. Enter your answer accurate to the fourth decimal place.

5. In a physics demo, Mr. Schmidgall applies a horizontal force to accelerate Mr. Smith (mass of 80 kg) rightward at a rate of 1.3 m/s/s. If the coefficient of friction between Mr. Smith's socks and the freshly waxed floors is .595, then with what force (in Newtons) must Mr. Schmidgall be pushing?

##### Solution Summary

Five good problems to learn Newton's laws of motion and motion of bodies are provided with detailed explanations of how to arrive at the answer.

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A 1-kg block of wood is dropped from a position 2.4 meters above a resting 0.05-kg ball. Shortly thereafter, the ball is shot horizontally at a velocity of 28 m/s. If the wood is 8.4 meters away from the ball in the horizontal direction, determine the distance (to two decimal places) the wood falls from its original position until impact? Ignore air resistance and use -9.8 m/s/s for the acceleration due to gravity.

Answer =

If the ball is to hit the wood both must be at the same point at the same time.
The time taken by the ball to cover the horizontal distance is given by t = 8.4/28 = 0.3 s.
As the initial vertical velocity of the ball is zero during this interval of time the vertical distance covered by the ball under gravity is given by
y = 0 + ½ g t2 = 0.5*9.8*0.09 = 0.441 m.
hence the distance the wood falls from its original position is h = 2.4 + y = 2.4 + 0.441 = 2.841 ...

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