mechanical problems and solutions

In practical, the particle does not change its state of rest or of uniform motion along a straight line unless it is forced to do this. This tendency of particle to do not change its state of rest or state of uniform motion along a straight line, unless that state is changed by an external force is called as inertia.

Mass is that quantity that is solely dependent upon the inertia of an object. The more inertia that an object has, the more mass that it has. Larger the mass of the particle, smaller will be the acceleration and hence larger will be the inertia.

The property which opposes the relative motion of the body over the surface of another body is called friction.

where μ is the coefficient of friction and N is the normal force.

Before going through any problem-related newton’s law of motion. You must have a strong hand over all the concepts related to it. Physics is a subject that helps us to understand the world. You should learn physics as you are helping yourself to understand how the different phenomenons happening in the world. The most inner core secret of Newton’s law of motion is the Free Body Diagram (FBD), this may help you to solve the problems very easily.

Example of free body diagram (FBD)

Sample Questions

Question 1: A passenger who is on a phone call while sitting on a train that is going at speed of 100 km/hr accidentally drops down his phone from the window. Neglecting air friction, what is the horizontal speed of the mobile phone just before it hits the ground?

Answer:

According to Newton’s first law of motion, object in a motion tends to stay in a motion unless until any external force is not acting. As there is no air friction acting on a object (mobile phone) to slow down the object in the horizontal direction after it drops from the train and acceleration due to gravity would only affect in the vertical direction. So, horizontal speed of the mobile phone just before hitting ground would be approximately 100 km/hr.

Question 2: What net force is required to keep a 1.5 kg ball moving with a constant velocity of 40 m/s?

According to Newton’s first law of motion,every body continues to be in its state of rest or of uniform motion in a straight line until unless any external force is not acting. If the net external force on a body is zero, its acceleration is zero.Hence force needed is also zero. Therefore 0 N net force required to keep ball moving with constant velocity of 40 m/s.

Question 3: A 2000 kg of the spaceship is moving in space with a constant velocity of 1200 m/s. What is a net force acting on the spaceship (there is no gravitational force acting on the spaceship).

Newton’s first law of motion states that object remains in a motion until unless any external force is not acting on a object.In a space there is vacuum and there is no external air resistance.Hence, spaceship will travel at constant velocity of 1200 m/s with zero acceleration. Since, m= mass of spaceship = 2000 kg a= acceleration of spaceship = 0 ∑F = m×a = 2000 × 0 = 0 N Hence, net force is acting on a spaceship is 0 N.

Question 4: What is meant by static and kinetic friction?

Resistance encountered by a body in static condition while tending to move under the action of an external force is called static friction. In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome.It is denoted as μ s . The resistance encountered by sliding body on a surface is known as kinetic friction. Kinetic friction is denoted as μ k . Kinetic friction is defined as a force that acts between moving surfaces. A body moving on the surface experiences a force in the opposite direction of its movement. The magnitude of the force will depend on the coefficient of kinetic friction between the two materials.

Question 5: If a car of mass 200 kg is moving with an acceleration of 5 m/s 2 , then what will be the net force of a car?

Given that, Mass of a car = M c = 200 kg Acceleration of a car = a c =5 m/s 2 Using formula F = M c × a = 200 × 5 = 1000 N Therefore, the net Force is 1000 N.

Question 6: A batter hits back a ball straight in the direction of the bowler with a velocity of 20 m/s and the initial velocity of the ball was 12 m/s.If the mass of the ball is 0.10kg. Determine the change in momentum on the ball.

Given that, Initial velocity of the ball = 12 m/s Final velocity of the ball = 20 m/s Mass of the ball = 0.10kg Change in momentum = final momentum – initial momentum = m×v2 – m×v1 = 0.10×20 – (-0.10×12) (ball again is in the direction from the batsman to the bowler) = 3.2 N.s Therefore, the change in momentum is 3.2 N.s.

Question 7: During training, a policeman fired a bullet from his gun on a wooden block, now a bullet of mass 10 gm is moving at 400 m/s penetrates 4 cm into a wooden block before coming to rest. Assuming that the force exerted by the wooden block is uniform, find the magnitude of force?

Given that, Mass of the bullet = M b = 10 gm = 0.010 kg Penetration of bullet before coming to rest = s = 4 cm = 0.04 m. Initial velocity of bullet = V i =400 m/s Final velocity of bullet = V f = 0 m/s Here, wooden block will exert force opposite in the direction of velocity,therefore this force causes deceleration. Hence a be the deceleration in this case (-a) By using kinematic equation, (V f ) 2 = (V i ) 2 + 2as ——(1) 0 = (400) 2 – 2 × a × 0.04 a = ( (400) 2 – 0 ) / 2 × 0.04 = 160000 / 0.08 = 2000000 The force on the bullet = M b × a = 0.01 × 2000000 = 20000 N

Question 8: A box of mass 100 kg is placed on a floor exerting some force on the floor. Determine what force does the floor exerting on the box? ( Here g= 9.81 m/s 2 ).

According to Newton’s third law motion,every action there is equal and opposite reaction.Hence the force exerted by the floor on the box will be the weight of box. Given that, Mass of box = M = 100 kg. weight of the box = M × g = 100 × 9.81 = 981 N The force exerted by the floor on the box = -981 N This Negative sign indicates that force applied by floor is in opposite direction of force applied by the box. Therefore, the Force applied by the floor is equal to 981 N.

Question 9: Define inertia of rest, motion , and direction?

A characteristic of matter that allows it to remain in its current condition of rest or uniform motion in a straight line until it is disrupted by an external force is called an inertia. Inertia of rest: The inability of a body to change its state of rest by itself is called inertia of rest Inertia of motion: The inability of a body to change its state of motion by itself is inertia of motion. Inertia of direction: The inability of a body to change its direction of motion by itself inertia of direction.

Question 10: There are two passengers in an elevator who have masses that exert a force of 180 N in the downward direction. They experience a normal force upwards from the elevator floor of 207 N.At what rate they are accelerating in the upward direction? (Here g=10 m/s 2 )

Given that, Upward force = 207 N Downward force = 180 N As they are accelerating in the upward direction then the net force- Net Force = ∑F = Upward force – Downward force = 207 -180 = 27 N To find total mass of the passengers, use the equation for force of gravity, F = m×g m = 180/10 m = 18 Kg To find net acceleration, use Newton’s second law of motion, F = m × a a = 27/18 a = 1.5 m/s 2 Therefore, they are accelerating in the upward direction at the rate of 1.5m/s 2 .

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Major American Universities Ph.D. Qualifying Questions and Solutions - Physics

Problems and solutions on mechanics.

The material for these volumes has been selected from the past twenty years' examination questions for graduate students at the University of California (Berkeley), Columbia University, the University of Chicago, MIT, State University of New York at Buffalo, Princeton University and the University of Wisconsin.

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PART II ANALYTICAL MECHANICS

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Index to Problems

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Collection of Solved Problems in Physics

Welcome in collection of solved problems in physics.

This collection of Solved Problems in Physics is developed by Department of Physics Education , Faculty of Mathematics and Physics , Charles University in Prague since 2006.

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The Collection contains tasks at various level in mechanics, electromagnetism, thermodynamics and optics. The tasks mostly do not contain the calculus (derivations and integrals), on the other hand they are not simple such as the use just one formula. Tasks have very detailed solution descriptions. Try to solve the task or at least some of its parts independently. You can use hints and task analysis (solution strategy).

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Mechanical Energy Problem Solutions

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Mechanical Energy Problems and Solutions

See examples of mechanical energy problems involving kinetic energy, potential energy, and the conservation of energy. Check your work with ours.

1. How much gravitational potential energy do you have when you lift a 15 N object 10 meters off the ground?

2. How much gravitational potential energy is in a 20 kg mass when 0.6 meters above the ground?

3. How much gravitational potential energy does a 35 kg boulder have when 30 meters off the ground?

4. How many times greater is an objects potential energy when three times higher?

If you need help on ratio problems click the link below:

Rule of Ones: analyzing equations to determine how other variables change

5. How much kinetic energy does a 0.15 kg ball thrown at 24 m/s have?

6. How many times greater is the kinetic energy of a ball that is going five times faster?

7. How much kinetic energy does a 1.2 kg ball have the moment it hits the ground 3.5 meters below when it starts from rest?

I cancelled out the initial kinetic energy because:

I cancelled out the final potential energy because:

8. How fast is a 1.2 kg ball traveling the moment it hits the ground 3.5 meters below when it starts from rest?

(Note: In many of these problems I could cancel out mass but did not since it was provided)

Since I did not cancel out mass I could answer the following questions if asked:

If I cancelled out mass in my work it would not show the actual initial potential energy since PE i = mgh and not just gh.

9. A 3.5 kg ball fell from a height of 12 meters. How fast is it traveling when its still 5 meters off the ground?

10. An 85kg roller coaster cart is traveling 4 m/s at the top of a hill 50 meters off the ground. How fast is it traveling at top of a second hill 20 meters off the ground?

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