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Akimi4 [234]
3 years ago
14

A 5 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with posit

ion as shown in the figure. The scale of the figure's vertical axis is set by Fs = 45 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?

Physics
1 answer:
jekas [21]3 years ago
5 0

The work done is 315 J

Explanation:

The graph is missing: find it in attachment.

The work done by a variable force is given by

W=\int F(x) dx

where

F(x) is the magnitude of the force

x is the position

On a force vs position graph, the work done is equivalent to the area under the graph. Therefore, in order to find the work done as the block moves from x = 0 to x = 8.0 m, we have to calculate the area under the graph between these two points.

Each square corresponds (vertically) to 45 N, so the area of the first trapezium between x = 0 and x = 4.0 m is

A_1 = \frac{(4+2)\cdot 2\cdot 45)}{2}=270

Then we have to find the area of the triangle between x = 6.0 m and x = 8.0 m:

A_2 = \frac{1}{2}(2)(45)=45

Therefore, the total area (and the work done) is

W=A_1+A_2=270+45=315 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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g You drop a 3.6-kg ball from a height of 3.5 m above one end of a uniform bar that pivots at its center. The bar has mass 9.9 k
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Answer:

h = 3.5 m

Explanation:

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2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

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vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

(2)(9.81\ m/s^2)(3.5\ m) = v_f^2 - (0\ m/s)^2\\v_f = \sqrt{68.67\ m^2/s^2}\\v_f = 8.3\ m/s

Now, we will apply the law of conservation of momentum:

m_1v_1 = m_2v_2

where,

m₁ = mass of colliding ball = 3.6 kg

m₂ = mass of ball on the other end = 3.6 kg

v₁ = vf = final velocity of ball while collision = 8.3 m/s

v₂ = vi = initial velocity of other end ball = ?

Therefore,

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Now, we again use the third equation of motion for the upward motion of the ball:

2gh = v_f^2 - v_i^2\\

where,

g = acceleration due to gravity = -9.81 m/s² (negative for upward motion)

h = height = ?

vf = final speed = 0 m/s

vi = initial speed = 8.3 m/s

Therefore,

(2)(9.81\ m/s^2)h = (0\ m/s)^2-(8.3\ m/s)^2\\

<u>h = 3.5 m</u>

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Answer:

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