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Energy calculation involving Work Done against friction

A box of mass 2 kg has an initial speed of 10 m/s at the foot of the ramp. Given that the friction along the ramp is 2 N, calculate the height h that it reaches when the speed of the box is 5 m/s. (g = 10 m/s2  work done inclined plane

Solutions:
For this question, the additional thing to note is the work done against friction. All the energy possesses by the box at bottom is KE. This KE will decrease and be converted to remaining KE at height h + gain in GPE + work done against friction.


To find the work done against friction, we need to find the distance moved by the box on the inclined ramp, d.

sin60 = h / d
d = h / sin60

Conservation of energy,
KE at bottom = KE at h + GPE at h + friction force

1/2mv2 = 1/2mv2 +   mgh  +   F x d
(1/2 x 2 x 102) = (1/2 x 2 x 52) + (2 x 10 x h) + (2 x h/sin60)
h     = 3.36 m

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Ratio of KE and GPE at various positions

A ball in thrown vertically up. D is the highest point the ball reached. Find the ratio of KE at B to the PE at C.   ratio of ke and gpe

Solutions: 1 : 1

To solve such question, you have to apply conservation of energy. In other words, total energy of ball at any positions (A, B, C and D) is the same (no air resistance etc). There is no way to find KE as no speed is given. So to find KE, you need to find it indirectly with the help of PE.

Understand the concept here as there can be many variations of question asked.


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Work Done, Energy and Power are Scalar Quantities

Work Done, Kinetic Energy (KE), Gravitational Potential Energy (GPE) and Power are all SCALAR quantities.

Many have asked if Work Done is scalar or vector? Though from our textbook, Work Done = Force x distance in the direction of the force.

In other words, Work Done = Force x Displacement

We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. In tertiary education, you will learn in details that the product (multiplication) of two vectors will result in a scalar or dot product.

Since it is out of syllabus, it will be good to use other ways to help you to understand and recall.

Method 1 Remember vector x vector will end up a scalar. Similar to (-)x(-) = (+) Method 2 Consider moving a box up to a certain height, work done or gain in GPE is constant regardless of the path it takes (diagonal or vertical). Hence direction is not important, therefore GPE  is a scalar.

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When considering KE, since KE = 1/2mv2 , similar to Method 1, v2 is actually velocity x velocity. Hence it product of 2 vectors, resulting in KE being a scalar.

Power = Energy Conversion/ time    or     Work Done / time,
since energy, work done and time are scalar, Power is a scalar quantity.

Other important note: Power = Rate of work done = Rate of energy conversion.


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Work Done, Energy and Power are Scalar Quantities

Work Done, Kinetic Energy (KE), Gravitational Potential Energy (GPE) and Power are all SCALAR quantities.

Many have asked if Work Done is scalar or vector? Though from our textbook, Work Done = Force x distance in the direction of the force.

In other words, Work Done = Force x Displacement

We already know that Force is a vector. Weight being a force, is also a vector quantity. Displacement is distance in a specific direction, hence it is a vector quantity too. In tertiary education, you will learn in details that the product (multiplication) of two vectors will result in a scalar or dot product.

Since it is out of syllabus, it will be good to use other ways to help you to understand and recall.

Method 1 Remember vector x vector will end up a scalar. Similar to (-)x(-) = (+) Method 2 Consider moving a box up to a certain height, work done or gain in GPE is constant regardless of the path it takes (diagonal or vertical). Hence direction is not important, therefore GPE  is a scalar.

Media_httpevantohfile_auomx

When considering KE, since KE = 1/2mv2 , similar to Method 1, v2 is actually velocity x velocity. Hence it product of 2 vectors, resulting in KE being a scalar.

Power = Energy Conversion/ time    or     Work Done / time,
since energy, work done and time are scalar, Power is a scalar quantity.

Other important note: Power = Rate of work done = Rate of energy conversion.


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N2008P1Q15 – Kinetic Model of Matter

Gas inside a cylinder is heated slowly to a higher temperature. The pressure inside the cylinder remains constant as the piston moves outwards.

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How do the speed of the gas molecules and their rate of collision with the piston compare with their initial values at the lower temperature?

Solutions:

In short:  Temperature increases, Kinetic Energy increases, Rate of Collision decreases, Average Force on wall increases, Pressure constant.

As temperature increases, the speed of molecules increases, the kinetic energy of the air molecules increases.

As piston is free to move, it will move to the right such that the pressure remains constant (equal to atmospheric pressure outside). As the piston moves to the right, the volume inside the piston increases

Surface area in which the air molecules collide increases.

The rate of collision decreases as the number of molecules remains constant. With higher KE of molecules, the molecules will collide the wall with greater force. Though rate of collision decreases, with each collision having greater impact force, the average force acting on the wall of piston increases.

Since P = F / A, with greater force F, over a bigger area A, the pressure P remains constant. (Compared with previously,  smaller F over smaller A, but P constant)  

Misconception: Many think that the rate of collision remains the same, which is wrong. Apparently the effect of volume increases is more significant, hence rate of collision decreases, even though they collide with greater impact force. Hence overall force on wall still increases. If the speed of the molecules increases but the pressure remains constant, then the molecules must collide less frequently. If the rate of collision stayed the same, the pressure would increase.  If the rate of collision increased, the pressure would increase even more.


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Work done – Man jumps from height 3 m, what is the force exerted by his legs?

A boy of mass 40 kg jumps from rest from a platform of height 3 m. He lands by bending his knees and stops his body in 0.5 s after landing. What is the force exerted by his legs? man jump

Solutions: 620 N

To solve this question, you need to apply conservation of energy (COE) to find the kinetic energy (KE) that the boy possesses just before he reaches the ground.
After which, you can solve the question using (1) Conservation of energy or (2) Kinematics to solve.

Using COE to find the KE just before he reaches the ground.

Method 1: Using Conservation of Energy (COE)

Method 2: Using Kinematics


You must know these 2 methods and always think along this 2 directions when dealing with such questions.