Physics

Forces, Energy and Machines

281 Questions

Forces, energy, and machines are fundamental physics concepts focusing on mechanics, work, power, and simple machines. These topics regularly appear in general science sections of competitive exams. Use these questions to practice calculating resultant forces, mechanical advantage, and work done in various scenarios.

Resultant force calculationsMechanical advantage of leversWork and power equationsApparent weight in elevatorsBending moments

Forces, Energy and Machines Questions

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A boy can exert a maximum effort of $250\  N$, so he uses an inclined plane to lift the load up. What should be the minimum length of the plank used by him to lift a mass of $ 50 \ kg$ to a height of $1 \ m$ ?

  1. $2$ $ m $
  2. $5$ $ m $
  3. $25$ $ m$
  4. $6$ $ m$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work done  $W = mgh$
$W = 50\times 10\times 1$  $N$
$W=500 N $
$W =$ Effort $\times$ distance
Distance$=500/250= 2m$

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A coolie uses a sloping wooden plank of length $2.0 m$ to push up a drum of mass $100 kg$ into the truck at a height $1.0 m$. What is the mechanical advantage of the sloping plank ?

  1. 6

  2. 5

  3. 2

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Length of the sloping wooden plank , $l = 2 m$
Vertical height , $h = 1 m$
We, know that for a inclined plane,
Mechanical Advantage $= \dfrac{Length \ of \ the \ incline}{vertical \ height}$
$\Rightarrow \dfrac{l}{h} = \dfrac{2}{1} = 2$

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A fire tongs has its arms $20cm$ long. It is used to lift a coal of weight $1.5kgf$ by applying an effort at a distance $15cm$ from the fulcrum. Find the mechanical advantage of fire tongs.

  1. $3$
  2. $0.50$
  3. $0.90$
  4. $0.65$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Effort Arm $ = 15 cm$
Load Arm $ = 20 - 15 = 5 cm$
Mechanical Advantage $= \dfrac{\text{Effort Arm}}{\text{Load Arm}} = \dfrac{15}{5} = 3$

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A fire tongs has its arms $20 \ cm$ long. It is used to lift a coal of weight $1.5 \ kgf$ by applying an effort at a distance $15 \ cm$ from the fulcrum. Find the effort needed.

  1. $0.2 kgf$
  2. $1.5 kgf$
  3. $2.5 kgf$
  4. $2.0 kgf$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$Load= 1.5kgf$
$Load \ arm= 20cm$
$Effort \ arm=15 cm$
$Effort \ = 20 \times 1.5/15= 2.0kgf$
The fulcrum in tongs is the joint end of the arms

Multiple choice physics lever common machines terms related to machines introduction to simple machines

A boy can exert a maximum force of $10$kg, i.e he cannot lift vertically a load of mass more than $10$kg. Now if he wants to raise a load of mass $20$kg on to a high eall, he can do it with the help of an inclined place making an angle $\theta$ with the horizontal. Find $\theta$.

  1. $60^o$
  2. $80^o$
  3. $30^o$
  4. $20^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$sin \theta=\dfrac{1}{M.A}=\dfrac{Effort}{Load}=\dfrac{10}{20}=\dfrac{1}{2}$
Thus, by placing a wooden plank at $\theta=30^o$ an angle equal to $30^o$ with the horizontal group can push the load of mass $20$Kg at any height by exerting a force(or effort) of $10$kg.

Multiple choice physics lever common machines terms related to machines introduction to simple machines

If a machine overcomes a load $L$ and the distance travelled by the load is $25 m$. Similarly, the effort applied in the machine is $E$ and the distance travelled by effort is $75 m$, and $'T'$ is the time taken,then velocity ratio is:

  1. $\dfrac{1}{3}$
  2. $\dfrac{2}{3}$
  3. $\dfrac{3}{1}$
  4. $1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The VR is define as ratio of distance travelled by effort:distance travelled by load

$VR=\dfrac{{E}{} _{d}}{{L}{} _{d}}$


$VR=\dfrac{25}{75}$

$VR=\dfrac{1}{3}$

option A is correct.
Multiple choice physics the essence of change different forms of energy energy conversions energy transformations and energy transfers

A person of weight 60 kg wants to loose 5 kg by going up and down 10m high stairs. Assume he burns twice as much fat while going up than going down. If 1 kg of fat is burnt on expending 7000 kcal. How many times must he go up and down to reduce his 7 kg weight? (Take $  g=10 \mathrm{ms}^{-2} )  $

  1. $ 1.8 \times 10^{3} $ times
  2. $ 2.4 \times 10^{3} $ times
  3. $ 1.7 \times 10^{3} $ times
  4. $ 2.1 \times 10^{3} $ times
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Energy used to go up $=mgh=60\times 10\times 10=6000\,J$

Energy used to come down $\dfrac{6000}{2}=3000\,J$

Energy used in one round trip $=9000\,J$

$1\,cal=4.5\,J$

$1\,J=\dfrac{1}{4.2\,cal}$

$9000\,J=\dfrac{9000}{4.2}=2142.85\,cal$

$7000\,kilo\,cal$ is required to burn $1\,kg$ mass

To reduce $5\,kg$ mass, energy required $=7000\times 5=35000\,kilo\,val$

Number of trip $=\dfrac{35000\times 1000}{2142.85}=1.7\times 10^{3}$
Multiple choice physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

Find the magnitude of the force on a charge of $12\mu C$ placed at point where the potential gradient has a magnitude of $6\times 10^{5}V\ m^{-1}$

  1. $5.20\ N$
  2. $7.20\ N$
  3. $6.20\ N$
  4. $8.20\ N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Potential gradient is nothing but the rate of change of electric potential with position and it is equal to electric field at that point.


$\dfrac{dV}{dl}=E$=electric field

$\implies E=6\times 10^5Vm^{-1}$

Force on charge $=F=qE=12\times 10^{-6}\times 6\times 10^5$

$\implies F=7.2N$

Answer-(B)

Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

A number of discs, each of momentum $M  kg  {m}/{s}$ are striking a wall at the rate of $n$ discs per minute. The force associated with these discs, in newtons, would be

  1. $\displaystyle\frac{Mn}{60}$
  2. $60 Mn$
  3. $\displaystyle\frac{M}{60n}$
  4. $\displaystyle\frac{n}{60M}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The force is $F=\dfrac{\Delta P}{dt}$

$momentum=M kg.m/s$
change in time=rate of dics striking wall 
$\Delta t=\dfrac{1}{n/minute}=\dfrac{60}{n}$ 
$F=\dfrac{\Delta P}{dt}=\dfrac{M}{60/n}=\dfrac{Mn}{60}N$

Multiple choice physics force and newton's laws of motion momentum (p) introduction to momentum linear momentum

A parachutist with total weight $75kg$ drops vertically onto a sandy ground with a speed of $2m{ s }^{ -1 }$ and comes to a halt over a distance of $0.25m$. The average force from the ground on her is close to

  1. $600N$
  2. $1200N$
  3. $1350N$
  4. $1950N$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Initial velocity while dropping on the ground u=2m/s.
Parachutist was brought to rest within s=0.25 m
Therefore retardation f : $u^2=2fs; \; f=\frac{u^2}{2s}=8m/s^2$
Averarage force from ground on her close is F=mf=600 N