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3 years ago

A helicopter pulls upward by means of a rope on a 250 kg crate to lift it UNIFORMLY. What is the net force on the crate?

Physics

1 answer:

Cloud [144]3 years ago

3 0

Answer:

The net force = 0

Explanation:

The given information includes;

The mass of the crate = 250 kg

The way the helicopter lifts the crate = Uniformly (constant rate (speed), no acceleration)

In order to pull the crate upwards, the helicopter has to provide a force equivalent to the weight of the crate keeping the helicopter on the ground.

The weight of the crate = The mass of the crate × The acceleration due gravity acting on the crate

The weight of the crate, $F_w$ ↓ = 250 kg × 9.81 m/s² = 2,452.5 N

The force the helicopter should provide to just lift the crate, $F_{(helicopter)}$ ↑ = The weight of the crate = 2,452.5 N

The net force, $F_{(net)}$ = $F_{(helicopter)}$ ↑ - $F_w$ ↓ = 2,452.5 N - 2,452.5 N = 0

The net force = 0.

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An electron travels west to east with a kinetic energy of 10 keV. The Earth's magnetic field in Pittsburgh is 19,911.5 nT in the

ch4aika [34]

Explanation:

It is given that,

Kinetic energy of the electron, $E_k=10\ keV=10^4\ eV=1.6\times 10^{-15}\ J$

Let the east direction is +x direction, north direction is +y direction and vertical direction is +z direction.

The magnetic field in north direction, $B_y=19911.5\ nT$

The magnetic field in west direction, $B_x=-3257.1\ nT$

The magnetic field in vertical direction, $B_z=48381.8 \ nT$

Magnetic field, $B=(-3257.1i+19911.5j+48381.8k)\ nT$

Firstly calculating the velocity of the electron using the kinetic energy formulas as :

$E_k=\dfrac{1}{2}mv^2$

$v=\sqrt{\dfrac{2E_k}{m}}$

$v=\sqrt{\dfrac{2\times 1.6\times 10^{-15}}{9.1\times 10^{-31}}}$

$v=5.92\times 10^7 i\ m/s$ (as it is moving from west to east)

The force acting on the charged particle in the magnetic field is given by :

$F=q(v\times B)$

$F=1.6\times 10^{-19}\times (5.92\times 10^7 i\times (-3257.1i+19911.5j+48381.8k)\times 10^{-9})$

Since, $i\times j=k\ \\j\times k=i\\k\times i=j$

And, $i\times i=j\times j=k\times k=0$

$F=1.6\times 10^{-19}\times [1178 k-2864.20j]$

$|F|=1.6\times 10^{-19}\times \sqrt{1178^2+2864.20^2}$

$F=4.95\times 10^{-16}\ N$

(b) Let a is the acceleration of the electron. It can be calculated as :

$a=\dfrac{F}{m}$

$a=\dfrac{4.95\times 10^{-16}}{9.1\times 10^{-31}}$

$a=5.43\times 10^{14}\ m/s^2$

Hence, this is the required solution.

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3 years ago

A 62kg box is lifted 12 meters off the ground. How much work is done?

Temka [501]

Answer: 7291.2 joules

Explanation:

Work is done when force is applied on an object over a distance.

Thus, Workdone = Force X distance

Since Distance moved by box = 12 metres

mass of box = 62kg

Acceleration due to gravity when box was lifted is represented by g = 9.8m/s^2

Recall that Force = Mass x acceleration due to gravity

i.e Force = 62kg x 9.8m/s^2

= 607.6 Newton

So, Workdone = Force X Distance

Workdone = 607.6 Newton X 12 metres

Workdone = 7291.2 joules

Thus, 7291.2 joules of work was done.

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IrinaK [193]

False. They are arranged in a structure called a crystal lattice

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A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c

Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 $m/ s^{2}$

$F - w = ma \\$

$F = w + ma$

$F = m ( a +g )$

$F = 75 ( 1.5 + 10 ) \\$

$F = 75 ( 11.5 )$

$F = 862.5 N$

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.

Learn more about speed here:-

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Physics students study a piano being pulled across a room on a rug. They know that when it is at rest, it experiences a gravitat

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