An 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of 30o with the horizontal. The force pushing the crate is parallel to the slope. If the speed of the crate increases at a rate of 1.5 m/s2, find the work done by the force. Group of answer choices 260 J 200 J 1 J 200 kJ 140 J

Answers

Answer 1

The work done by the force is 140 J.

Calculation:

It is given that, an 80-N crate is pushed a distance of 5.0 m upward along a smooth incline that makes an angle of [tex]30^\circ[/tex] with the horizontal. The force pushing the crate is parallel to the slope and the acceleration is 1.5 [tex]\text{ m/s}^2[/tex].

It is required to find the work done by the force.

The free-body diagram is shown below,

Here, W=80 N, a=-1.5 [tex]\text{ m/s}^2[/tex]

We can write the net force equation as,

[tex]\begin{gathered}F-W \sin 30^{\circ}=m a \\F-m g \sin 30^{\circ}=\left(\frac{W}{g}\right) a\end{gathered}[/tex]

Here W is the weight of the object, m is the mass of the body, and F is the applied force.

Thus,

[tex]\begin{aligned}F-(80 \mathrm{~N}) \sin 30^{\circ} &=\left(\frac{80 \mathrm{~N}}{9.8 \mathrm{~m} / \mathrm{s}^{2}}\right)\left(-1.5 \mathrm{~m} / \mathrm{s}^{2}\right) \\F &=27.755 \mathrm{~N}\end{aligned}[/tex]

It is known that the work done by the force is calculated as the multiplication of the applied force and the displacement.

Thus the work one is,

[tex]\begin{aligned}W &=(27.755 \mathrm{~N})(5 \mathrm{~m}) \\& \approx 140 \mathrm{~J}\end{aligned}[/tex]

Thus the last option is the correct one.

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An 80-N Crate Is Pushed A Distance Of 5.0 M Upward Along A Smooth Incline That Makes An Angle Of 30o

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How can you use the period of a wave to find the frequency?

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Explanation:

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http://www.sengpielaudio.com/calculator-period.htm

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Answer:

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Explanation:

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Answers

Answer:

Net force = 5N

Explanation:

Jay = 20N to the right

Bradley = 15N to the left

To find the net force;

Since the forces are being applied to the box in opposite direction i.e acting in opposite direction, we would subtract them.

Net force = 20 - 15

Net force = 5N

Therefore, the net force on the box is 5 Newton.

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Answer:

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Answer:

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Explanation

The diagram in A. contain a variable resistor which can be use to vary the pd from the 6v battery

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Answer:

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Explanation:

Given

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Important Tip:

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Using the formula

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Explanation:

For this exercise let's use the relationship between momentum and momentum variation

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Answers

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Answers

Answer:

The electrostatic force that will result if both charges are doubled and the distance remains the same is 6.0 * 10⁻⁶ N

Explanation:

Coulomb's law of  electricity states that the magnitude of the force of attraction or repulsion between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance of of separation between them.

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The given data in the question ,

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F₁ is the electric force for case 1= 3.0 x 10⁻⁶ N

F₂ is the  electric force for case 2= ?

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[tex]\rm F_1=\frac{Kq_1q_2}{d_1^2}[/tex]

For case 2,

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[tex]\rm F_2=4\times\frac{Kq_1q_2}{d_1^2}\\\\\rm F_2=4F_1[/tex]

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Answer:

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Explanation:

Hope this helps :))

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