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3 years ago

2.

Physics

1 answer:

Alex_Xolod [135]3 years ago

5 0

Answer:

The work done on the sled by friction, W = - 4593.75 J

Explanation:

Given data,

The combined mass of sled and the boy, m = 75 kg

The displacement of the boy, S = 25 m

The coefficient of the friction, u = 0.25

The frictional force acting on the boy,

F = u η

Where,

η - is the normal force acting on the boy (mg)

Substituting the values,

F = 0.25 x 75 x 9.8

= 183.75 N

Since the direction of the frictional force is against the direction of motion

F = - 183.75 N

The work done on the sled by friction,

W = F x S

= - 183.75 x 25

= - 4593.75 J

Hence, the work done on the sled by friction, W = - 4593.75 J

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Answer:

$4.125\times 10^{-11}\ W/m^2$

Explanation:

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Intensity of light is given by

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4 years ago

A 1kg mass is thrown to a height of 2cm. what is the potential energy

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2 years ago

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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a

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Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.

The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.

Reasons:

The energy given to the block by the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

According to the principle of conservation of energy, we have;

On a flat plane, energy given to the block = $0.5 \cdot k \cdot x^2$ = kinetic energy of

block = $0.5 \cdot m \cdot v^2$

Therefore;

0.5·k·x² = 0.5·m·v²

Which gives;

x² ∝ v²

x ∝ v

On a plane inclined at an angle θ, we have;

The energy of the spring = $\mathbf{0.5 \cdot k \cdot x^2}$

The force of the weight of the block on the string, $F = m \cdot g \cdot sin(\theta)$

The energy given to the block = $0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)$ = The kinetic energy of block as it leaves the spring = $\mathbf{0.5 \cdot m \cdot v^2}$

Which gives;

$0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2$

Which is of the form;

a·x² - b = c·v²

a·x² + c·v² = b

Where;

a, b, and c are constants

The graph of the equation a·x² + c·v² = b is an ellipse

Therefore;

As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.

Please find attached a drawing related to the question obtained from a similar question online

The possible question options are;

As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x

The relationship is no longer linear and v will be more for the same value of x

The relationship is still linear, with lesser value of v

The relationship is still linear, with higher value of v

The relationship is still linear, but vary inversely, such that as x increases, v decreases

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