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# Category Archives: Mechanics

## Mechanics Questions from Mohammed

a. Determine the work done by the motor when the wheelchair starts at rest and speeds up to its normal speed.

b. Determine the maximum distance that the wheelchair can travel on a horizontal surface at its normal speed, using its stored energy. (Ignore the energy needed for it to speed up when it starts.)

c. Suppose that 0.023 percent of the power required for driving is expended against drag due to the flexing of the wheelchair’s soft rubber tires. Calculate the magnitude of the drag force.

## The Physics of Accidents

Why are road accident at high speeds very much worse than accidents at low speeds?

Asked Muneem.

Answer:

As you might have studied, greater the speed, greater is the momentum. When an accidents occurs, the greater momentum involved can cause greater damage.

Please ferer to the links below for details

1. http://outreach.phas.ubc.ca/phys420/p420_96/danny/danweb.htm

2. http://ffden-2.phys.uaf.edu/211_fall2002.web.dir/ben_townsend/PhysicsofCarCollisions.htm

## Why is it necessary to bend knees while jumping from a height?

Why is it necessary to bend knees while jumping from a height?

Asked Rahul. (via SMS from our mobile version http://m.askphysics.com )

Answer:

This is a consequence of the impulse momentum principle.

When we bent our knees on landing, the time of impact increases and reduces the average force exerted during the entire process. This prevents the chance of damage to our body.

Impulse momentum principle states that the impulse of a force is equal to the change in momentum produced.

Impulse is measured as the product of force and time interval. So, when we stop ourselves, there is a fixed change in momentum. If we stop ourselves very fast, the time interval is less and the average force will be very large.

But, if we bent our knees during landing, the time involved in the proces of landing will be more and thus the average force will be less and hence the damage will also be less.

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## What is centripetal force?

What is centripetal force?

Received via SMS from our mobile version of this site http://m.askphysics.com

Answer

Centripetal fforce is the net force to be acting on a body so that it can move in a curved path. The centripetal force is always directed towArds the center of the circular path. The word centripetal means “directed towards the center” or “center seeking force”.

Please remember that it is not a new kind of force, but in different case one or more of the fundamental forces or their components provide the necessary centripetal for e.

The magnitude of the centrifugal force that must be acting on a body of mass m moving with a speed v along a curve of radius of curvature r is given by F = mv^2/r

## Buoyancy

I read in one of my chapters that when a body is immersed in a container of water (kept on a weighing machine which reads ‘W’ at first )the apparent weight lost by the body is equal to the buoyant force, and thus the reading ‘W’ increases.

Then I came across a question like this –

A beaker containing water kept on a weighing machine weighs W. A body of weight ‘w’ is dropped in it. It is floating & experiencing a buoyant force B, then the reading on machine is –

(a)w+W

(b)W+B

(C)W+w-B

(d)W

please give me the answer and explanation

the book says answer is (a)

but I think it is (c)

(Anwesha posted this questions)

###### Answer

The answer is A.

The total force acting downwards is W+w and the weighing scale is providing an equal reaction. the normal reaction offered by the weighing scale is what we get as the reading.

therefore the reading on the weighing scale must be **W+w**

The buoyant force is acting on the object dropped and is not contributing to the normal reaction offered by the weighing scale.

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## Physics Numerical Problems posted by Nabeela

The following questions were posted by Nabeela. We are just posting the question as such so that the visitors can post the answers.

- A mass-less string pulls a mass of 20kg upward against gravity. The string would break if subjected to a tension greater than 400N. What is the maximum acceleration with which the mass can be moved upward?
- A bullet of mass 20g is fired from a gun of mass 10kg with a velocity of 180m/s. Find the velocity of recoil of the gun. Find the force required to stop the gun before it moves 20cm
- A body of mass 40kg is moving up an inclined plane with a uniform velocity when a force of 460N is applied. If the plane is inclined to the horizontal by an angle of 45o,calculate the coefficient of kinetic friction between the surfaces.
- If the coefficient of friction between the tyres of a truck and the road is k, show that the maximum stopping distance of the truck when moving with a velocity ‘v’ is v
^{2}/2 kg. Assume that the brakes are not applied - A ball of mass 0.5kg moving with a speed of 20m/s collides with an identical ball at rest. After collision the direction of each ball makes an angle of 30o with the original direction. Find the speed of each ball after collision.
- What is the maximum horizontal distance that a ball thrown with a speed of 60 m/s can go without hitting the roof of a long hall 30m high?
- A ball is thrown horizontally strikes a wall 5m away. The height of the point struck by the ball is 1m lower than the height which it was thrown from. (1) With what velocity was it thrown (2) At what angle did the ball reach the wall?
- The time of flight of aprojectile is 10 seconds . its range on a horizontal plane is 100m . Calculate the angle of projection and the velocity of projection.

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## Four persons K, L , M, N are initially at the four corners of a square of side d Problem

Four persons K, L , M, N are initially at the four corners of a square of side ‘d’. Each person now starts moving with a uniform speed ‘v’ in such a way that K always moves directly towards L, L towards M, M towards N and N towards K. After what time will they meet?

Gayathri asked.

**Answer:** Many students do not understand the real situation initially. Every time the persons are approaching each other and hence they will be moving closer and closer as they continue their walking. Finally they’ll reach the centre of the square to meet each other. I’ve tried to visualize the situation below.

From the diagram, we can make out that the resultant displacement by each when they meet will be d/√2 and the component of velocity of each towards the final point (the centre of the square) is v/√2.

Therefore,the time taken = displacement by the component of velocity in its direction = (d/√2) / (v/√2) = **d/v**

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