# Newton's Laws of Motion: 5 Problems

See the attached file.

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?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Please see the attachment.

55305

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 + Â½ 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 m

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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.

Answer =

The acceleration 'a' of the puck can be determined using first equation of motion

v = u + at

where v is the final velocity, u is the initial velocity and t is the time interval.

Substituting the values we get

5.67 = 11.5 + a*7.45

gives a = (5.76 - 11.5)/7.45 or a = - 0.782 m/s/s -ve sign shows that it is retardation and is due to the force of friction.

If the mass of the puck is m and Î¼ is the coefficient of friction, then the friction force is - Î¼mg and using law of motion (F = ma) we can write

- Î¼ mg = m (-0.782) gives Î¼ = 0.782/9.8 = 0.07979.

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

Answer =

The distance traveled by the puck during this interval is given by the second equation of motion, that is

V2 = u2 + 2ax

Or 5.762 = 11.52 + 2*(-0.782)*x gives x = 63.345 m.

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.

Answer =

If Î¼ is the coefficient of kinetic friction and m is the nass of the bobsled then the friction force is F = - Î¼mg and the acceleration is a = F/m = - Î¼g = - 9.8*Î¼

Using equation of motion

V2 = u2 + 2ax we get

0 = 30.52 + 2*(- 9.8*Î¼)*144 gives

Î¼ = 930.25/2822.4 = 0.32959

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.

Answer =

As above the acceleration due to friction is

- Î¼ g = - 5.17 or Î¼ = 5.17/9.8 = 0.52755

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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?

Answer =

The net force on Mr. Smith is F - Î¼ mg, where F is the force applied by Mr. Schmidgall, so using the laws of motion (F = ma) we get

F - Î¼mg = ma

Or F = ma + Î¼mg = m(a + Î¼g) = 80*(1.3 + 0.595*9.8) = 570.48 N.

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