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Alchen [17]
3 years ago
5

A 2kg box is pushed along a flat frictionless surface with an applied force of 53.91newton j+19.62 j were j is horizontal and j

is vertical if the box moves 10m horizontal in 90 seconds what is the work and power that is required to complete the movement
Physics
1 answer:
tamaranim1 [39]3 years ago
5 0

Since the box doesn't move vertically at all, no work is done by the vertical component of the force.

If 53.91 Newtons is the horizontal component of the force (very unclear in the question), then the work done is

<u>Work = (force) x (distance)</u>

Work = (53.91 N) x (10 m)

<em>Work = 539.1 Joules</em>

<u>Power = (work done) / (time to do the work)</u>

Power = (539.1 joules) / (90 sec)

Power = (539.1/90) (joule/sec)

<em>Power = 5.99 watts</em>

=====================================

Note:  If the mass of the box is only 2 kg, and you push it along the surface with a constant force of 53⁺ Newtons, and the surface really is frictionless, then that box is gonna cover a WHOOOOOLE LOT more than 10 meters in 90 seconds.  I get 109,168 meters !

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Explanation:

Full solution calculation can be found in the attachment below.

From the principle of conservation of linear momentum, the sum of momentum before collision equals the sum of momentum after collision.

Before collision only the train had momentum. After the collision the train and the boxcars stick together and move as one body. The initial momentum of the train is now shared with the boxcars as they move together as one body. The both move with a common velocity v.

See the attachment below for the solution calculation.

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Hello, the tripping of a 230-kilovolt transmission line.

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How many doctors are ther in Edmonton​
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Consider a space shuttle which has a mass of about 1.0 x 105 kg and circles the Earth at an altitude of about 200.0 km. Calculat
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Substitute into the formula

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g = 6.67×10^-6/4×10^10

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The amount of gas that a helicopter uses is directly proportional to the number of hours spent flying. the helicopter flies for
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Then,

\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

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\therefore \frac{g_1}{g_2}=\frac{T_1}{T_2}

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