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 accelerated motion calculus methods of motion equations equation of motion the equation of motion and its derivation

A ladder of length 10 m and mass 20 kg (and with uniform mass distribution) leans against a slippery vertical wall. The ladder makes an angle of $30^{\circ}$ with respect to the vertical. Friction between the ladder and the ground prevents it from sliding downwards. What is the magnitude of the force exerted on the ladder by the wall?
$[Take \sqrt { 3 } =1.732;g=10{ m/s }^{ 2 }]$

  1. $0 N$
  2. $0.58 N$
  3. $58 N$
  4. $5.8 N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using torque equilibrium about the base: (Weight * L/2 * sin(30)) = (Force_wall * L * cos(30)). Thus Force_wall = (Weight/2) * tan(30). Weight = 20kg * 10m/s^2 = 200N. Force_wall = 100 * (1/sqrt(3)) = 100 / 1.732 = 57.73 N, which rounds to 58 N.

Multiple choice physics accelerated motion calculus methods of motion equations equation of motion the equation of motion and its derivation

At an airport, a bored child starts to walk backwards on a moving platform. The child accelerates relative to the platform with $a =  - 0.5m/{s^2}$ relative to the platform. The platform moves with a constant speed $v =  + 1.0m/s$ relative to the stationary floor. In $4.0$ seconds, how much will the child have been displaced relative to the floor?

  1. $8m$
  2. $0m$
  3. $3m$
  4. $4m$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
initial vel = vel of platform $=1\ m/s$
acceleration wrt floor $=$ acceleration of him wrt platform +acc. of platform
$=\dfrac {-1}{2}m/s^2+0=\dfrac {-1}{2}m/s^2$
So, for finding displacement of him in us,
Applying second equation of motion
$x=ut +\dfrac {at^2}{2}$
$(1)(4)^2-\left (\dfrac {1}{4}\right) \dfrac {(4)^2}{2}$
$x=4-\dfrac {(4)^2}{4}=0\ m$

Multiple choice physics accelerated motion calculus methods of motion equations equation of motion the equation of motion and its derivation

An elevator car whose floor-to-ceiling distance is equal to 2.7m starts ascending with a constant acceleration $1.2m/{s^2};2.0s$ after the start a bolt begins falling from the ceiling of the car. Find the displacement covered by the bolt during the free fall in the reference frame fixed to the elevator shaft.

  1. 0.7 m

  2. 1.7 m

  3. 2.7 m

  4. 3.7 m
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the total time of free fall be $t$
Then at $t=2s$
$4 _{B}=2. \dfrac{1}{m}/s$          $4 _{C}=2.\dfrac{1}{m}/s$
$a _{C}=1.2\ m/s^{2}$              $a _{B}=-10\ m/s^{2}$
Wrt car.
distance travelled by bolt $= 27\ m$
Then 
$-2.7\ m=(4 _{B/C})t+a _{B/C}\left(\dfrac{t^{2}}{2}\right)$
$\Rightarrow -2.7= (2.4-2.4)t (\dfrac{-10-12}{2}) t^{2}$
$\Rightarrow 2.7= \dfrac{11.2}{2}t^{2}$
$\Rightarrow t^{2}=\left(\dfrac{2.7}{11.2}\times 2\right)^{-11}=0.482\ s^{2}$
Note: $4 _{B}=4 _{C}$ till $t=2.3$ before the both spots to fall. 
$4 _{B}=4 _{C}=a _{C}t=(1.2)(2)=2.4 m/s$
let the displacement of the bolt be $h$ Then, 
$h=4 _{B}(t)+4 _{\dfrac{B}{2}}t^{2}$
$h=(2.4)\sqrt{0.482}\dfrac{-(10)}{2}(0.482)$
$h=-0.743$
$h=\approx -0.7\ m$
$|h|\approx 0.7\ m$
Option $A$ is correct

Multiple choice physics accelerated motion calculus methods of motion equations equation of motion the equation of motion and its derivation

A cart begins from rest at the top of a long incline and rolls with a constant acceleration of $2m/s^{2}$. How far has the cart moved along the incline after rolling for $3$ seconds?

  1. $3$ meters
  2. $6$ meters
  3. $9$ meters
  4. $18$ meters
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial velocity of cart      $u =0  m/s$

Acceleration of the cart       $a = 2$ $m/s^2$
Distance covered by cart in 3 seconds       $S = ut + \dfrac{1}{2}at^2$                  where  $t =3$ seconds
$\therefore$    $S = 0 + \dfrac{1}{2 } \times 2 \times 3^2  = 9$  meters

Multiple choice forces on solids elastic and plastic substances forces and matter properties of matter physics

An external force of $10\ N$ acts normally on a square area of each side $50\ cm$. The stress produced in equilibrium state is

  1. $10\ N/m^{2}$
  2. $20\ N/m^{2}$
  3. $40\ N/m^{2}$
  4. $50\ N/m^{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Stress = Force / Area. The force is 10 N. The side of the square is 50 cm = 0.5 m, so the area is 0.5 * 0.5 = 0.25 m^2. Stress = 10 / 0.25 = 40 N/m^2.

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A man carries a load on his head through a distance of 5 m. The maximum amount of work is done when he

  1. Movies it over an inclined plane

  2. Movies it over a horizontal surface

  3. Lift it vertically upwards

  4. All of the above
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The maximum work done by man will be when he lift it vertically upwards because in such situation the man has to exert force opposite to gravity that is in the direction of the displacement of load. 

Multiple choice physics energy and its forms introduction to work work introduction to work and energy

A force of $5 N$ is applied on a $20 kg$ mass at rest. the work done in the third second is:-

  1. $\dfrac{25}{8}J$
  2. $\dfrac{25}{4}J$
  3. $12 J$
  4. $25 J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Displacement in third second = displacement till 3rd second - Displacement till 2nd second 

=$\dfrac{a}{2}3^2-\dfrac{a}{2}2^2=\dfrac{5}{8}m$ (No term of ut because u=0)

Thus work done in third second =$5\times\dfrac{5}{8}=\dfrac{25}{8}J$

Multiple choice maths how big? how heavy? measuring volume volume of solids volume of cube and cuboid

A beam 4m long, 50cm wide and 20cm deep is made of wood which weighs 25$kg$ per $m^3$. Find the weight of the beam.

  1. $10 kg$
  2. $12 kg$
  3. $13 kg$
  4. $15 kg$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given,length $=4m$
breadth $=50 cm=0.5 m$
height $20 cm=0.2 m$
Volume of beam $=l \times b \times h=0.4 m^3$
Weight of $1$ $m^3$ wood is $25$ kg
Weight of $0.4$  $m^3$ cubiodal beam$=25\times 0.4=10 $kg

Multiple choice physics turning on a pivot the turning effect of a force moment of force or torque turning effect of force couple

A body is acted upon by two forces each of magnitude $F$, but in opposite directions. State the effect of the forces if the two forces act at two different points of the body at a separation $r$. Which of the following is/are true?

  1. Resultant force $=0$
  2. moment of forces $=Fr$

  3. The forces tend to rotate the body about the mid-point between the two forces.

  4. All of the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Here we apply two forces of same magnitude but opposite in direction.Hence we have F and - F which makes resultant force as zero.


Moment of force for F is $r \times F $. For force F it has distance r therefore it has moment rF and same with force -F.Hence moment of force is rF.

The forces tend to rotate the body about the mid-point between the two forces.

Multiple choice physics sources of energy biomass and biogas biogas conventional sources of energy

Wind with a velocity $72 km/hr$ blows normally against one of the walls of a house with an area of $100 m$ (density of air is $1.2 kgm^{-2}$)

  1. Force exerted on the wall if the air moves parallel to wall after striking is $48000 N$
  2. Force exerted on the wall if the air moves normal to the wall after striking is $96000 N$

  3. Force exerted on the wall is maximum if air moves parallel to the wall after striking

  4. Force exerted on the wall is maximum if air moves at certain angle with wall after striking
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics rotational motion of a rigid body and moment of inertia constant angular acceleration equation of motion of rotating body dynamics of rotational motion about a fixed axis

Newton's second law of motion and work done in rotation of a rigid body can be expressed as 

  1. Newton's law cannot be expressed in rotation, work done in rotation is $W=\tau \ theta$
  2. Force and work done are expressed as $\tau = I \alpha$ and $W=\tau \ theta$
  3. Force can be expressed as $\tau = I \alpha$, while work done will be zero

  4. Force will be zero, since no net displacement is present
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Newton's second law of motion in kinematics is F =  ma. To express the same in rotation, replace mass by moment of inertia I AND linear acceleration a by angular acceleration $\alpha$. Thus, Newton's law of motion becomes, Torque $\tau = I \alpha$

Similarly work done in kinematics is given by W = F.S. To express the same in rotation, replace Force by torque $\tau$ AND linear displacement by angular displacement $\theta$. Thus, Work done becomes, Torque $W= \tau \theta$

Multiple choice physics force the turning of couple couple turning effect of force

Two small kids weighing 10 kg and 15 kg are trying to balance a seesaw of total length 5m, with the fulcrum at the centre. If one of the kids is sitting at an end, where should the other sit?

  1. $2.5 m$
  2. $1 m$
  3. $1.7 m$
  4. $2 m$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a seesaw to balance, the torque on both sides must be equal. Let the fulcrum be at 0. One kid (10 kg) is at 2.5 m from the center. The other kid (15 kg) must be at distance x such that 10 * 2.5 = 15 * x. Solving for x gives 25 / 15 = 1.666... m, which is approximately 1.7 m.

Multiple choice physics free, damped and forced oscillations forced vibration forced vibrations free, forced and damped oscillations

A force F= -4x-8 is acting on a block where x is position of block in meter. The energy of oscillation is 32 J, the block oscillate between two points Position of extreme position is:

  1. 6

  2. 0

  3. 4

  4. 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$F=-4x-8\F=-4(x+2)\ \Rightarrow F=-4[x-(-2)]\F=-kx\rightarrow$ Distance mean position. So, this will be SHM

with $x=-2$ as mean position 
as $K _SHM=4$
Energy of SHM$=\cfrac{1}{2}KA^2=32\ \Rightarrow A^2=\cfrac{2\times32}{K}=\cfrac{2\times32}{4}=16\Rightarrow A=4$
and mean position $=-2$
Extereme position $=(-2+4)\quad and (-2-4)\=2\quad and\quad -6$