Tag: weightlessness

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Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

The backside of a truck is open and a box of 40kg is placed 5m away from the rear end.The coefficient of friction of the box with the surface of the truck is 0.15.The truck starts from rest with $2m/s^2$ acceleration.Calculate the distance covered by the truck when the box falls off

  1. 20m

  2. 30m

  3. 40m

  4. 50m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The box experiences a pseudo-force F = m*a = 40 * 2 = 80N. The friction force is f = mu * m * g = 0.15 * 40 * 10 = 60N. The net force on the box is F_net = F - f = 80 - 60 = 20N. Acceleration of the box relative to the truck is a_rel = F_net / m = 20 / 40 = 0.5 m/s^2. Time to fall off: s = 0.5 * a_rel * t^2 => 5 = 0.5 * 0.5 * t^2 => 5 = 0.25 * t^2 => t^2 = 20 => t = sqrt(20). Distance covered by truck: S = 0.5 * a_truck * t^2 = 0.5 * 2 * 20 = 20m.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A jet water issues from a nozzel with a velocity of $20 m/s$ and it impinges normally on a flat plate moving away from it at $10 m/s$. If the cross-sectional area of the jet is $0.02 m^2$ and the density of water is taken as $1000 kg/m^3$, then the force developed on the plate will be

  1. $ 10 N$
  2. $ 100 N$
  3. $1000 N$
  4. $2000 N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

A jet water issues from a nozzel with a velocity$=20m/s$

It impinges normally on a flat plate moving away from it$=10m/s$
Cross-sectional area of the jet $=0.02m^2$
The density of water is taken$=1000kg/m^3$
The force developed on the plate will be
Hydrostatic force o the bottom surface of the tank will be
$F _b=\rho g(l\times b)\times h\F _b\rho g\times(1\times2)\times 2\ F _b=4\rho g(lbh)\rightarrow(1)$
Hydrostatic force on vertical surface will be
$F _v=\rho g(l\times h)\times\cfrac{h}{2}\F _v=\rho g(2\times 2)\cfrac{lh^2}{2}\F _v=2\rho g(lh^2)\rightarrow(2)$
Ratio will be
$\cfrac{F _b}{F _V}=\cfrac{4\rho g(lbh)}{2\rho g(lh^2)}\ \cfrac{F _b}{F _V}=1\F _b=1\times10\ \quad=10N$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A man drops an apple in the lift. He finds that the apple remains stationary and does not fall. The lift is:

  1. Going down with constant speed

  2. Going up with constant speed

  3. Going down with constant acceleration

  4. Going up with constant acceleration
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As the apple is dropped, it is under free-fall meaning that the force of gravity is acting on it. With respect to the person inside the lift, the apple seems not to be falling Hence, the man and the lift must also be falling with the action of acceleration due to gravity i.e, a constant acceleration.

option - C is correct.

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

When a lift is going up with uniform acceleration, the apparent weight of a person is $W _{1}$
When the lift is stationary, his apparent weight is $W _{2}$
When the lift falls freely his apparent weight is $W _{3}$
When the lift is going down with uniform acceleration which is less than the acceleration due to gravity, his apparent weight is $W _{4}$
The increasing order of these four weights is

  1. $W _{1},W _{3},W _{2},W _{4}$
  2. $W _{3},W _{4},W _{2},W _{1}$
  3. $W _{3},W _{2},W _{4},W _{1}$
  4. $W _{2},W _{3},W _{4},W _{1}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. $W _{1}>mg$
When lift is stationary, $W _{2}=mg$
When lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. $W _{3}=0$
When lift accelerates down, pseudo force acts upwards,  hence it decreases apparent weight $W _{4}<mg$, but also the acceleration is less than g, therefore $W _{4}=m(g-a)>0$

Hence, $W _{3}<W _{4}<W _{2}<W _{1}$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

Two identical trains A and B move with equal speeds on parallel tracks along the equator. A moves from east to west and B moves from west to east. Which train exerts greater force on the track?

  1. A

  2. B

  3. they will exert equal force

  4. The mass and the speed of each train must be known to reach a conclusion.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Working in the frame of universe:

$mg-N=m{ \omega  }^{ 2 }R$
$N=mg-m{ \omega  }^{ 2 }R$
Earth moves from west to east.
Now, since the train B moves from west to east, its $\omega$ is greater than A. 
Hence, ${ N } _{ B }<{ N } _{ A }$

Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A lift of mass $1000 \ Kg$ which is moving with acceleration of $1 m/s^{-2}$ in upward direction, then the tension developed in string which is connected to lift is : -

  1. $9,800 N$
  2. $10,800 N$
  3. $11,000 N$
  4. $10,000 N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The mass of the lift is $1000\;{\rm{kg}}$, the acceleration is $1\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$ and the acceleration due to gravity is $9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}$.

The tension developed in string is given as,

$T = mg + ma$

$ = 1000 \times 9.8 + 1000 \times 1$

$ = 10800\;{\rm{N}}$

Thus, the tension developed in the string is $10800\;{\rm{N}}$.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

WEIGHTLESSNESS
An astronaut experiences weightlessness in a space satellite. It is because

  1. the gravitational force is small at that location in space.

  2. the gravitational force is large at that location in space.

  3. the astronaut experiences no gravity.

  4. the gravitational force is infinitely large at that location in space.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An astronaut experiences weightlessness in a space satellite. It is because the astronaut experiences no gravity.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

A spaceship is launched into a circular orbit close to earth's surface. The additional velocity that should be imparted to the spaceship in the orbit to overcome the gravitational pull is:
(Radius of earth $=6400km$ and $g=9.8m\quad { s }^{ -1 }$)

  1. $11.2km\quad { s }^{ -1 }$
  2. $8km\quad { s }^{ -1 }$
  3. $3.2km\quad { s }^{ -1 }$
  4. $1.5km\quad { s }^{ -1 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,
Radius of the earth, $R=6400km$
Acceleration due to gravity, $g=9.8m/s^2$
When a satellite is orbitting the earth its velocity is its orbital velocity given by orbital velocity: 
 ${ V } _{ 0 }=\sqrt { \dfrac { GM }{ R }  } =\sqrt { g } R$    $[\because g =\frac{GM}{R^2}]$
If it has to overcome gravitational pull then its velocity should be escape velocity escape velocity ${ V } _{ e }=\sqrt { \dfrac { 2GM }{ r }  } =\sqrt { 2gR } $
Additional velocity required is $V={ V } _{ e }-{ V } _{ 0 }=\sqrt { 2gR } -\sqrt { gR } =\sqrt { gR } \left( \sqrt { 2 } -1 \right) $
given that $R=64004m=64\times { 10 }^{ 6 }m$
$g=9.8m/{ s }^{ 2 }$
So, additional velocity will be:
$V=\sqrt { gR } \left( \sqrt { 2 } -1 \right) =\sqrt { 9.8\times 6.4\times { 10 }^{ 6 } } \left( \sqrt { 2 } -1 \right) $
$\Rightarrow V=3.28\times { 10 }^{ 3 }m/s$
Hence, the correct option is $(C)$
Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

Astronauts on the orbiting space station are weightless because...

  1. there is no gravity in space and they do not weigh anything.

  2. space is a vacuum and there is no gravity in a vacuum.

  3. space is a vacuum and there is no air resistance in a vacuum.

  4. None of the reasons given above are correct
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As the astronauts are in a frame which is rotating around the earth and the centripetal acceleration on astronauts and the satellite is $same$, $g=GM/r^2$ where $r$ is the radius of the orbit. So relative to the satellite the astronauts can't exert any force on satellite so no reaction force from the satellites so they feel the condition of weightlessness.


Its same like an elevator falling from top to ground freely $cable$ broken.

So the man in the elevator and the elevator have same acceleration $g$ so the man can't exert any force on the elevator so the condition of weightlessness.

Option D is correct.

Multiple choice physics gravitation: planets and satellites weightlessness application of newton's law of motion escape velocity

In which of the following situations, you will not experience weightlessness?

  1. If you are at the center of the earth

  2. You are falling freely under gravity of the earth

  3. Inside a spacecraft which is in an orbit around the earth

  4. In a lift moving downwards with uniform velocity
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Weightlessness occurs when the normal force is zero. At the center of the Earth, gravitational force is zero, but you are not in free fall in the same sense as an orbit; however, you would experience zero weight. Wait, actually, in a lift moving with uniform velocity, the acceleration is zero, so the normal force equals your weight (mg), meaning you feel your normal weight. Therefore, you do NOT experience weightlessness in a lift moving with uniform velocity.