Physics

Forces, Energy and Machines

267 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 simple machine common machines terms related to machines introduction to simple machines

In a lifting machine, an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10,000 N by a distance of 0.8 m. Determine the velocity ratio and mechanical advantage.

  1. 25 and 20

  2. 23 and 22

  3. 20 and 30

  4. 25 and 35
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Distance moved by effort is 20m,that of load is 0.8m

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

$VR=\dfrac{20}{0.8}$

$VR=25$

the load is 10,000N and effort=500N

$MA=\dfrac{{load}{} _{d}}{{effort}{} _{d}}$

$MA=\dfrac{10000}{500}$

$MA=20$
Multiple choice physics simple machine common machines terms related to machines introduction to simple machines

In a lifting machine, an effort of 500 N is to be moved by a distance of 20 m to raise a load of 10,000 N by a distance of 0.8 m. Determine the effort lost in friction

  1. 100 N

  2. 120 N

  3. 80 N

  4. 0 N
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Work input = 500 N * 20 m = 10,000 J. Work output = 10,000 N * 0.8 m = 8,000 J. Work lost to friction = 10,000 - 8,000 = 2,000 J. Effort lost = 2,000 J / 20 m = 100 N.

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

It is easier to draw up a wooden block along an inclined plane than to haul it vertically, principally because:

  1. the friction is reduced

  2. the mass becomes smaller

  3. only a part of the weight has to be overcome
  4. $g$ becomes smaller
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vertical plane $W=mgh$
for inclined plane $W'=mghsin(Q)$
$sinQ<1$
$W>W'$
hence only one part of weight is to overcome

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

Akhil has to lift a load of 800 N onto a platform of 2 m height. Instead, he pushes the load up a ramp 4 m long. Find the force required to roll the load up the ramp.

  1. 200 N

  2. 600 N

  3. 400 N

  4. 800 N
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Force required to roll the load up the ramp (F) $=\displaystyle \frac{load \times height}{distance}$
$= 800 \displaystyle \times \frac{2}{4} = 400 N$

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

Akhil has to lift a load of 800 N onto a platform of 2 m height. Instead, he pushes the load up a ramp 4 m long. Find the work done by Akhil in lifting the load vertically up.

  1. 800 J

  2. 1600 J

  3. 400 J

  4. 200 J
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The work done by Akhil in lifting the load vertically up $=$ Force $\times$ height
$= 800 \times 2 = 1600 \ J$

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

A nut can be opened by a lever of length $0.25$ m by applying a force of $80$ N. What should be the length of the lever if a force of $32$ N is enough to open the nut?

  1. $625$ m
  2. $1625$ m
  3. $0.625$ m
  4. $6.25$ m
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the principle of moments: F1 * L1 = F2 * L2. 80 N * 0.25 m = 32 N * L2. L2 = (80 * 0.25) / 32 = 20 / 32 = 0.625 m.

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

A man of mass $80kg$ carrying a load of $20kg$ walks up a stair case in $20s$. If the number of steps is $40$ and width and height of each step are $20cm$ and $15cm$ respectively. The efficiency of the man is

  1. $20 \%$
  2. $25\%$
  3. $40\%$
  4. $50\%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total mass = 100 kg. Total height = 40 * 0.15 m = 6 m. Work done = mgh = 100 * 10 * 6 = 6000 J. Power output = 6000 / 20 = 300 W. Assuming human efficiency is related to power, this question is ambiguous without a defined input power.

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

A man uses a crowbar of length $1.5m$ to raise a load of $75kgf$ by putting a sharp edge below the bar at a distance $1m$ from his hand. Calculate the mechanical advantage. 

  1. $0$
  2. $1$
  3. $2$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Load $ = 75 kgf$
Length of crowbar $= 1.5 m$
Effort arm $= 1 m $
So, Load Arm $= \text{Length of crowbar - Effort arm}$


$\Rightarrow 1.5 - 1 = 0.5 m$
Mechanical Advantage $= \dfrac{\text{Effort Arm}}{\text{Load Arm}} = \dfrac{1}{0.5} =2$

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

A man uses a crowbar of length $1.5m$ to raise a load of $75kgf$ by putting a sharp edge below the bar at a distance $1m$ from his hand. Calculate the effort needed.

  1. $75kgf$
  2. $375kgf$
  3. $37.5 gf$
  4. $37.5kgf$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Load $ = 75 kgf$
Length of crowbar $= 1.5 m$
Effort arm $= 1 m $
So, Load Arm $= \text{Length of crowbar - Effort arm}$
$\Rightarrow 1.5 - 1 = 0.5 m$
Mechanical Advantage $= \dfrac{\text{Effort Arm}}{\text{Load Arm}} = \dfrac{1}{0.5} =2$


Effort needed $= \dfrac{\text{Load}}{\text{Mechanical Advantage}} = \dfrac{75}{2} = 37.5 kgf$

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