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# Tag Archives: Newton’s laws of motion

## Physics Video Project

The video project was given in 2008 for XI class students of Meghalaya. The best video among them is posted here. Though there are some errors, the video is inspiring to note the work done entirely from scratch by the XI Class students.

## Numericals from Newton’s Laws of motion based on F=ma (For calss XI students of Kendriya Vidyalaya Pattom)

Download the questions and solve them and submit on or before 9 Oct 2014

1. A force acts for 10 s on a body of mass 10 kg after which the force ceases and the body describes 50 m in the next 5 s. Find the magnitude of the force. [Ans: 10 N]
2. A  truck starts from rest and rolls down a hill with constant acceleration. It travels a distance od 400 m in 20s. Calculate the acceleration and the force acting on it if its mass is 7 metric tonnes. [Ans: 2 m/s2, 14000N]
3. A motor car running aat the rate of 7 m/s can be stopped by applying brakes in 10 m. Show that total resistance to the motion, when the brakes are on is one fourth of the weight of the car.
4. In an Xray machine, an electron is subjected top a force of 10-23 N. In how much time the electron will cover a distance of 0.1 m,? Take the mass of the electron = 10-30 kg

## Numericals from Newton’s Laws of motion based on F=ma (For calss XI students of Kendriya Vidyalaya Pattom)

Download the questions and solve them and submit on or before 9 Oct 2014

1. A force acts for 10 s on a body of mass 10 kg after which the force ceases and the body describes 50 m in the next 5 s. Find the magnitude of the force. [Ans: 10 N]
2. A  truck starts from rest and rolls down a hill with constant acceleration. It travels a distance od 400 m in 20s. Calculate the acceleration and the force acting on it if its mass is 7 metric tonnes. [Ans: 2 m/s2, 14000N]
3. A motor car running aat the rate of 7 m/s can be stopped by applying brakes in 10 m. Show that total resistance to the motion, when the brakes are on is one fourth of the weight of the car.
4. In an Xray machine, an electron is subjected top a force of 10-23 N. In how much time the electron will cover a distance of 0.1 m,? Take the mass of the electron = 10-30 kg

## Force and acceleration – Can force be exerted without acceleration?

I am not able to understand this and I couldn’t think of a better example. i repeat. its just an example. if i punch someone’s face with my arm having constant velocity, it will mean that the acceleration of my arm is zero. F=ma, therefore force exerted by my arm is zero. Now if the force applied is zero the why does the person whom i punched experiences pain….. please help… thanks… It’s not whether your hand is moving with constant velocity, bu what happens to your hand after hitting the face of the opponent. The face resists the motion of your hand the momentum of your hand is imparted/shared to the face. And there is (-)acceleration to the hand and causes a force to be exerted on the face. The momentum imparted to the face will be responsible for the damage caused. ## Does Newton’s Third Law apply on throwing butter on a wall?

Newton’s third law of motion states that

“When we throw any object on anything, that thing also pushes back with the same force. But when we throw butter on the wall, it sticks on the wall. Why?”

Asked by Akshit and Koushal from Class VII A from Kendriya Vidyalaya Air Force Station Bidar, Karnataka

I feel that there is some misconception/misunderstanding here. Newton’s Third law states that whenever two bodies interact each other, the force exerted by one body on the other is equal and opposite to the force exerted by the second body on first. These forces are simultaneous and are exerted on different bodies. tha is; the force exerted by the first body is acting on the second body and the force exerted by the second body is acted on the first body.

It doesn’t matter whether the body sticks to the other body or bounces back. We are concerned here with force and not with the motion.

When handful of butter (or clay) is thrown on to the wall, the ball of butter exerts some force on the wall. The wall exerts an equal force simultaneously. But the adhesive force between the wall and butter is more than the force acting on butter which tries to detach it. So it is not detached. When a rubber ball thrown on the wall, the force exerted by the wall on the ball compresses it. Due to the elastic nature of the ball, it tries to bring the shape back to original and this makes it bounce back. The force exerted on the wall due to the bouncing ball is double that of a clay/butter ball of same mass hitting the the wall with same speed; because, the rubber ball requires the force to bounce back too.

http://www.phy-astr.gsu.edu/dhamala/Physics2211/Chapter9.pdf ## More questions waiting for answers:

• When a piece of paper is held with its face perpendicular to a uniform electric field the flux through it is 25.0 N·m2/C. When the paper is turned 25.0o with respect to the field the flux through it is?
• A cylinder rests on a horizontial rotating  disc, find at what angular velocity the axes of the disc and cylinder is R and the coefficient of friction U>D/H, where D is the diameter of the cylinder and H is its height.
• Can static friction do non zero work on an object?if yes give example . if no give reason
• How angular momentum is different from linear momentum? ## Body at incline 1. In first case we have the box with weight, say, 10 Newtons on incline. We calculate two components (perpendicular and parallel)of weight. We calculate normal force and force of friction. The box is at rest.

2. In second case we have ball on incline, made of same material as box from first case. Weight of the ball is the same as in first case, components of weight are the same as in first case. Normal force and friction have the same amount as in the case of box. But,unlike the box, ball is moving – it slides down. How’s that possible? If calculated net force is zero?

It is not true that the ball slides,  but it will roll down.

In the first case, the forces balance each other and there is no motion. In the second force, the frictional force acting tangentially backwards (up the incline) and the component of weight of the ball acting parallel to the plane and through the centre of the ball constitute a couple and tends to rotate it. Now there is no sliding; it rolls.

(If any further clarification is required please post as comment to this post)

Please refer to http://www.real-world-physics-problems.com/rolling-without-slipping.html for detailed treatment of the Physics of rolling without friction. ### Hits so far @ AskPhysics

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