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3 years ago

Work done can be describe as?

Physics

1 answer:

Neko [114]3 years ago

4 0

Explanation:

In physics, work is the energy transferred to or from an object via the application of force along a displacement. In its simplest form, it is often represented as the product of force and displacement. A force is said to do positive work if (when applied) it has a component in the direction of the displacement of the point of application. A force does negative work if it has a component opposite to the direction of the displacement at the point of application of the force.

Quick Facts: Common symbols, SI unit ...

Work

A baseball pitcher does positive work on the ball by applying a force to it over the distance it moves while in his grip.

Common symbols

SI unit

joule (J)

Other units

Foot-pound, Erg

In SI base units

1 kg⋅m2⋅s−2

Derivations from

other quantities

W = F ⋅ s

W = τ θ

Dimension

M L2 T−2

For example, when a ball is held above the ground and then dropped, the work done by the gravitational force on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force F is constant and the angle between the force and the displacement s is θ, then the work done is given by:

{\displaystyle W=Fs\cos {\theta }}{\displaystyle W=Fs\cos {\theta }}

Work is a scalar quantity, so it has only magnitude and no direction. Work transfers energy from one place to another, or one form to another. The SI unit of work is the joule (J), the same unit as for energy.

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1) given that the only force acting on the ball is gravity, gravity acts along the vertical axis. Since no other force acts on the ball then the horizontal velocity will remain constant all through the flight since there is no horizontal force acting on the ball.

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Answer:

5080.86m

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$v_1=v_{01}+a_1t$

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We calculate then the height achieved in part 1:

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And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

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$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

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$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

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Work done can be describe as?​

Work done can be describe as?