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 energy : forms and sources concept of energy energy for everything forms of energy

A man carries a heavy box on his head on a horizontal plane form one place to another. In this he does

  1. Maximum work

  2. No work

  3. Negative work

  4. Minimum work
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Work is defined as force multiplied by displacement in the direction of the force. When carrying a load horizontally, the force (gravity) acts vertically, while the displacement is horizontal. Since the angle between force and displacement is 90 degrees, the work done is zero.

Multiple choice maths surface area and volume of cube and cuboid length of the diagonal diagonal of cube and cuboid surface area of cubes and cuboids

A guy wire attached to a vertical pole of height $18m$ is $24m$ long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

  1. $15.87m$
  2. $16.8m$
  3. $15$
  4. $15.67$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider the wire and pole system as a right angled triangle such that length of pole is the perpendicular, length of wire the hypotenuse and distance between the base of pole and the stack is the base, such that, H = 24 m, P = 18 m
Now, from Pythagoras theorem,
$H^2 = P^2 +B^2$
$24^2 = 18^2 + B^2$
$576 = 324 + B^2$
$B^2 = 252$
$B = 15.87$ m
Thus, distance between the base of the pole and the stack is 15.87 m

Multiple choice physics trigonometrical ratios angles and sides naming the sides in a right angled triangle angle and their measurement

If the distance between a 13-foot ladder and a vertical wall is $5$ feet along the ground, how high can a person climb if the ladder is inclined against wall?

  1. $18$ feet
  2. $65$ feet
  3. $\cfrac{13}{5}$ feet
  4. $8$ feet
  5. $12$ feet
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Length of Ladder = 13 ft

Length between Ladder and Wall = 5 ft
by pythagoras theorum
(hypotenuse)^2 = (perpendicular)^2 + (base)^2
(13)^2 = (height of wall)^2+(5)^2
by solving
Height of Wall = 12 ft

Multiple choice physics force and types of force introduction to force what is force? concept of force and its unit

If force $\vec {F} = 5\hat {i} + 3\hat {kj} + 4\hat {k}$ makes a displacement of $\vec {s} = 6\hat {i} - 5\hat {k}$, work done by the force is

  1. $10\ unit$
  2. $122\sqrt {5}\ unit$
  3. $5\sqrt {122}\ unit$
  4. $20\ unit$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that 

Work done $=force \times  displacement$
Work done $=\left( {5\widehat i + 6\widehat j + 4\widehat k} \right) \cdot\left( {6\widehat i + 0\widehat j - 5\widehat k} \right)$
Work done$=30-20=10$ $unit$
Hence,
option $(A)$ is correct answer.

Multiple choice dielectric substances and polarization dielectrics and polarisation electrostatic potential and capacitance electrostatics physics

Force $'F'$ exists between two point charges in air, if a medium having $\epsilon _{R} = 6$ is inserted between charges, the new force between them.

  1. Will increase by $6$ times
  2. Will decrease by $6$ times
  3. Will decrease by $\sqrt {6}$ times
  4. Will increase by $36$ times
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The force in a medium is F_medium = F_air / K. With K=6, the force decreases by 6 times.

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

A 50.0 kg boy is sitting on an amusement park ride where he accelerates straight upward from rest to a speed 30.0 m/s in 3.0 s. What is his mass as he accelerates upward?

  1. 990.0 kg

  2. 100.0 kg

  3. 50.0 kg

  4. 5.00 kg

  5. 0 kg
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Mass of an object always remains constant whether the object is accelerating or not. Apparent weight of the object changes due to acceleration.

Hence the mass of the boy is $50$ kg even he accelerates upward.
Thus option C is correct.

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

A plumb bob is hung from the ceiling of a train compartment. the train moves on an inclined track of inclination $30^\circ $ with horizntal. The acceleration of train up the plane is $a=\,g/2$. The angle which the string supporting the bob makes with normal to the ceiling in equilibrium is-

  1. $30^\circ $
  2. ${\tan ^{ - 1}}\left( {2/\sqrt 3 } \right)$
  3. ${\tan ^{ - 1}}\left( {\sqrt 3/2 } \right)$
  4. ${\tan ^{ - 1}}\left( 2 \right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Multiple choice physics newton's laws of motion weightlessness application of newton's law of motion escape velocity

A man of mass 'm' stands on a weighing machine in a lift

List - I  List-II
(a) Lift moves up with uniform acceleration a  (d) mg
(b) Lift moves down with uniform acceleration a (e) m(g$+$a)
(c) Lift moves down with uniform velocity (f) m(g-a)
  1. $a\rightarrow e,b\rightarrow f,c\rightarrow d,$
  2. $a\rightarrow d,b\rightarrow f,c\rightarrow e,$
  3. $a\rightarrow d,b\rightarrow e,c\rightarrow f,$
  4. $a\rightarrow f,b\rightarrow d,c\rightarrow e,$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When the lift moves $UP$ with uniform acceleration $a$ $Pseudo$ force on the man is $ma$ downward.

Net downward force is $mg+ma$. Hence, $e$

When the lift moves $down$ with uniform acceleration $a$ $Pseudo$ force on the man is $ma$ upward.
Net downward force is $mg+ma$. Hence, $f$


When the lift moves down with $uniform \ velocity$, only force acting is gravity.

Hence net force on man is $mg$. Hence, $d$

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

If the position of two like parallel forces shifted by one-fourth of the distance between the forces when the two forces are interchanged. The ratio of the two forces is:

  1. $1:2$
  2. $2:3$
  3. $3:4$
  4. $3:5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let forces be F1 and F2 at distance d. The resultant acts at x = F2*d / (F1+F2). If forces are swapped, new position x' = F1*d / (F1+F2). The shift is |x - x'| = d * |F1-F2| / (F1+F2) = d/4. Thus, 4|F1-F2| = F1+F2. If F1 > F2, 4F1 - 4F2 = F1 + F2 => 3F1 = 5F2 => F1/F2 = 5/3. If F2 > F1, 4F2 - 4F1 = F1 + F2 => 3F2 = 5F1 => F1/F2 = 3/5. Option D matches 3:5.

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 energy and its forms idea of energy introduction to work and energy work and energy

The force acting on a 4gm mass in the energy region ${ U=8x }^{ 2 }$ at x= -2 cm is :

  1. 8 dyne

  2. 4 dyne

  3. 16 dyne

  4. 32 dyne
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given,

$m=4gm$
$U=8x^2$

The force acting on mass $m$ in the potential energy $U$  at $x=-2cm$ will be

$F=-\dfrac{dU}{dx}$
$F=-\dfrac{d(8x^2)}{dx}$
$F=-16x$
$|F| _{x=-2cm}=-16\times(-2)=32dyne$
The correct option is D.



Multiple choice physics pressure in liquids and gases common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

A barometer kept in an elevator reads $76\ cm$ when it is at rest. If the elevator goes up with some acceleration, the reading will be

  1. $76\ cm$
  2. $> 76\ cm$
  3. $< 76\ cm$

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

A barometer measures atmospheric pressure based on the weight of the mercury column. When an elevator accelerates upward, the effective gravity (g_eff = g + a) increases, which would normally increase the pressure reading; however, the mercury column itself also experiences this increased effective gravity, causing the height to remain constant at 76 cm.