A Team Of Eight Dogs Pulls A Sled With Waxed Wood Runners ✓ Solved

A Team Of Eight Dogs Pulls A Sled With Waxed Wood Runners

Problem 1 (3 points): A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 19.0 kg, and the loaded sled with its rider has a mass of 210 kg. (a) Calculate the acceleration starting from rest if each dog exerts an average force of 185 N backward on the snow. (b) What is the acceleration once the sled starts to move? (c) For both situations, calculate the force in the coupling between the dogs and the sled. (coeff of Static friction = 0.14, coeff of kinetic friction = 0.1) Problem 2 (4 points): At takeoff, a commercial jet has a 60.0 m/s speed. Its tires have a diameter of 0.850 m. (a) At how many rev/min are the tires rotating? (b) What is the centripetal acceleration at the edge of the tire? (c) With what force must a determined 1.00x 10-15kg bacterium cling to the rim? (d) Take the ratio of this force to the bacterium’s weight. Problem 3 (4 points): Riders in an amusement park ride shaped like a Viking ship hung from a large pivot are rotated back and forth like a rigid pendulum. Sometime near the middle of the ride, the ship is momentarily motionless at the top of its circular arc. The ship then swings down under the influence of gravity. (a) Assuming negligible friction, find the speed of the riders at the bottom of its arc, given the system’s center of mass travels in an arc having a radius of 14.0 m and the riders are near the center of mass. (b) What is the centripetal acceleration at the bottom of the arc? (c) Draw a free body diagram of the forces acting on a rider at the bottom of the arc. (d) Find the force exerted by the ride on a 60.0 kg rider and compare it to her weight.

Paper For Above Instructions

Introduction

This paper addresses several problems related to forces and motion, focusing on the application of Newton's laws, friction, rotational motion, and centripetal forces. Each problem will be analyzed step-by-step to find the required solutions.

Problem 1: Sled Dogs and Friction

(a) The total mass of the dogs is:

Mass of dogs = number of dogs × average mass per dog

Mass of dogs = 8 × 19.0 kg = 152 kg

Total mass (sled + dogs) = 210 kg + 152 kg = 362 kg

Each dog exerts a force backward, and the total force exerted by all the dogs is:

Total force = 8 × 185 N = 1480 N

The frictional force (static, as the sled starts moving) can be calculated as:

Frictional force (static) = coefficient of static friction × Normal force

Normal force = total mass × g = 362 kg × 9.81 m/s² = 3557.22 N

Frictional force (static) = 0.14 × 3557.22 N = 497.99 N

The net force acting on the sled when starting from rest is:

Net force = Total force - Frictional force = 1480 N - 497.99 N = 982.01 N

The acceleration can be calculated using Newton's second law (F = ma):

Acceleration = Net force / Total mass = 982.01 N / 362 kg = 2.72 m/s²

(b) Once the sled starts moving, kinetic friction takes effect. The frictional force (kinetic) is calculated as:

Frictional force (kinetic) = coefficient of kinetic friction × Normal force = 0.1 × 3557.22 N = 355.72 N

The new net force is:

Net force = 1480 N - 355.72 N = 1124.28 N

Thus, the acceleration while the sled is moving becomes:

Acceleration = 1124.28 N / 362 kg = 3.10 m/s²

(c) The force in the coupling can be calculated as the force exerted by the dogs minus the frictional force:

Force in coupling (from rest) = 1480 N - 497.99 N = 982.01 N

Force in coupling (moving) = 1480 N - 355.72 N = 1124.28 N

Problem 2: Jet Tire Rotation

(a) To calculate the rotational speed of the tires, we first find the circumference:

Circumference = π × diameter = π × 0.850 m = 2.67 m

The tires complete one revolution per circumference. Therefore, the number of revolutions per second can be calculated as:

Revolutions per second = speed / circumference = 60.0 m/s / 2.67 m = 22.49 rev/s

In revolutions per minute (rev/min):

Revolutions per minute = 22.49 rev/s × 60 s/min = 1349.4 rev/min

(b) The centripetal acceleration at the edge of the tire can be calculated as:

Centripetal acceleration = v² / r, where v is the speed and r is the radius of the tire:

Radius = diameter / 2 = 0.850 m / 2 = 0.425 m

Centripetal acceleration = (60.0 m/s)² / 0.425 m = 8461.18 m/s²

(c) The force required for the bacterium can be calculated using centripetal force:

Centripetal force = mass × centripetal acceleration = 1.00 × 10^-15 kg × 8461.18 m/s² = 8.46 × 10^-12 N

(d) The weight of the bacterium is calculated as:

Weight = mass × g = 1.00 × 10^-15 kg × 9.81 m/s² = 9.81 × 10^-15 N

The ratio of the force to the weight is:

Ratio = Force / Weight = (8.46 × 10^-12 N) / (9.81 × 10^-15 N) = 861.67

Problem 3: Amusement Park Ride

(a) The speed of the riders at the bottom can be found using energy conservation as the potential energy at the top converts to kinetic energy at the bottom:

Potential Energy (PE) = mgh, and Kinetic Energy (KE) = 0.5mv²

mgh = 0.5mv²

g = 9.81 m/s² and h = 14.0 m; thus:

v = √(2gh) = √(2 × 9.81 m/s² × 14.0 m) = 16.76 m/s

(b) The centripetal acceleration at the bottom is given by:

Centripetal acceleration = v² / r = (16.76 m/s)² / 14.0 m = 19.95 m/s²

(c) A free body diagram includes the gravitational force acting downward (weight) and the normal force acting upward from the ride.

(d) The force exerted by the ride on a 60.0 kg rider at the bottom can be given by the normal force, which combines the gravitational force and the centripetal force:

Normal Force = Weight + Centripetal Force

Weight = mg = 60.0 kg × 9.81 m/s² = 588.6 N

Centripetal Force = m × centripetal acceleration = 60.0 kg × 19.95 m/s² = 1197 N

Total Force = 588.6 N + 1197 N = 1785.6 N

This force is significantly greater than the rider's weight, emphasizing the effect of centripetal force.

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