IRSA IMPORTANT QUESTIONS

Question 1

A seesaw is balanced with a pivot at its centre. On one side, there is a child weighing 40 kg sitting 2 meters away from the pivot. How far from the pivot should another child, weighing 30 kg, sit on the opposite side to keep the seesaw in balance?

Answer 1

2.67

Question 2

A solid beam 0.5 m long is laid across a pivot to form a simple lever. The pivot is 0.1 m from the end of the beam. Calculate the heaviest load that could be lifted using a force of 500 N.

Answer 2

Moments must first be taken, then using the moment calculate we must find the maximum load from pivot 0.1

Question 3

How can a lever be used to lift heavy objects?
How does the length of the lever arm affect the moment?

Answer 3

A lever can be used to lift heavy objects by placing the load closer to the fulcrum and applying the effort force at a greater distance. This arrangement increases the mechanical advantage, making it easier to lift the object.

The longer the lever arm, the greater the moment. Increasing the distance from the fulcrum increases the lever’s mechanical advantage.

Question 4

Why is it difficult to open a door when you push it near the hinges?

Answer 4

When you push a door near the hinges, the moment arm (distance from the pivot point to the point where the force is applied) is small. According to the moment formula (moment = force × distance), a smaller moment arm requires a larger force to create the same moment. Hence, it’s difficult to open the door because you need to apply more force to overcome the small moment arm.

Question 5

Why do see-saws balance when children of different weights sit on opposite ends?

Answer 5

A see-saw balances because of moments. When a heavier child sits closer to the pivot and a lighter child sits farther away, the product of their weight and distance from the pivot (moment) on both sides remains equal. This balanced moment keeps the see-saw stable, demonstrating the principle of rotational equilibrium.

Question 6

Why do engineers consider moments when designing structures like bridges?

Answer 6

Engineers consider moments in bridge design to ensure stability and safety. Bridges experience various forces, including the weight of the bridge itself and the loads from vehicles and pedestrians. By calculating and distributing these loads to create balanced moments, engineers ensure that the bridge structure remains stable and can withstand different forces without collapsing. Understanding moments is essential for designing structurally sound and secure bridges

Question 7

Why doesn’t a book placed on a table accelerate downwards, even though gravity is pulling it down?

Answer 7

The book doesn’t accelerate downwards because the force of gravity is balanced by the normal force exerted by the table upward. When an object is at rest on a surface, like the book on the table, the forces are in equilibrium, meaning the forces are balanced, and there is no net force acting on the object.

Question 8

Find the acceleration of the object if the force applied is 250 N and the mass of the object is 50 kg.

Answer 8

5

Question 9

Find the mass of the object if the force applied is 225N and the acceleration of the object is 15 m/𝑠^2.

Answer 9

15

Question 10

A passenger in an elevator has a mass that exerts a force of 110N downwards. He experiences a normal force upwards from the elevator’s floor of 130N. What direction is he accelerating in, if at all, and at what rate? Use g=10 m/𝑠^2

Answer 10

Calculate net force and note that he is decelerating.

Using F=mg you must calculate the mass of the passenger, m = f/g 110/10 = 11kg

Then using the mass and net force you can calculate the acceleration using F=ma where a = m/f

11/20 = 1.81ms

Question 11

5
A car of mass =1500 kg is moving with a constant velocity of 20 m/son a straight, level road. Suddenly, the driver applies the brakes, bringing the car to a stop in 10 s.
Calculate:
The force exerted by the brakes to stop the car.
Explain how this problem illustrates Newton’s first law of motion.

Answer 11

a = change in /time = 20/10 = 2

f=ma 1500 x 2 = 3000

Question 12

Scenario: A box of mass 20 kg is placed on an inclined plane with an angle of 30 degrees. If the coefficient of friction between the box and the plane is 0.4, calculate the force of friction acting on the box when it’s on the verge of sliding down.

Answer 12

using fnet = fric x fnormal

fnormal = w=mg x cos 30

20 x 9.81 x cos 30 x fric(0.4) = 31.92N

Question 13

Scenario: A 1500 kg car accelerates from rest to a speed of 25 m/s in 10 seconds. Calculate the average force exerted by the car during this acceleration.

Answer 13

a = change in v/t = 25/10 = 2.5

1500 x 2.5 = 3750N

Question 14

: A spacecraft of mass 2×10^5 kg is moving through space at a constant velocity of 3×10^4 m/s. Suddenly, its engines fail, and no external forces act on it. Calculate the time it takes for the spacecraft to come to a complete stop.

Answer 14

The spaceship will continue moving at constant velocity indefinitely unless acted upon by an external force, as the external force is zero as stated by newtons first law. if v = 0 and u = 3x10^4 then a = 0 t can be calculated as t = v-u/a when the values are plugged the answer is undefined

Question 15

Scenario: A block of mass 4 kg is placed on a frictionless inclined plane with an angle of 30∘ to the horizontal. Calculate the acceleration of the block down the incline.

Answer 15

F down mgsin30 = 4 x 9.81 x sin 30 = 19.62

19.62 x 0.5 =9.8N

using f down =ma then 9.8/4 = 2.45

Question 16

What force is friction?

Answer 16

Friction is a contact force which acts only when two objects are in contact with each other. It is an opposing force i.e. it always opposes the motion of the object.

Question 17

Which is the weakest force in nature?

Answer 17

Among all the fundamental forces “Gravitational Force” is the weakest force and is also considered to be the weakest force in nature.

Question 18

What is the difference between Newton’s First Law and Inertia?

Answer 18

Inertia is the property of any object that it acquires either in a state of rest or in a state of motion. Newton’s first law is a law which states that an object always maintains its Inertia of rest or Inertia of motion.

Question 19

Why do objects slow down?

Answer 19

In general, an object moving with a constant velocity slows down because of friction

Question 20

A 1500 kg spaceship travels in the vacuum of space at a constant speed of 600 m/s. Ignoring any gravitational forces, what is the net force on the spaceship?

Answer 20

In a vacuum, there is no friction due to air resistance. Newton’s first law states that an object in motion stays in motion unless acted upon by a net force. Thus, the spaceship will travel at the constant speed of 600 m/s and the net force on the spaceship must be zero as acceleration is also zero.

Since,

F = ma

Therefore,

F = (1500kg)(0m/s2)
= 0 N.

Question 21

A ball rolls off the back of a train going 50 mph(mile per hour). Neglecting air friction, what is the horizontal speed of the ball just before it hits the ground?

Answer 21

Newtons first law states than an object in motion tends to stay in motion unless acted upon by an external force.
Neglecting air friction, there is no external force to slow the ball down in the horizontal direction after it falls off the train.
The acceleration of gravity would only affect the ball in the vertical direction.
So, the horizontal speed of the ball is 50 mph.

Question 22

Answer the following question
1.Briely explain the role of Physics and its Relation to other discipline.
2.Write down the SI unit for mass, time and temperature
3.Write down the derived unit for acceleration, Velocity and Power
4.In order to calculate force what two more quantities are required?
5.The time for 10 swings of a pendulum is 13 +/- 0.3s, what is the time and uncertainty for 1 swing? #

A 20 kg object is placed on a flat horizontal surface. Calculate the gravitational force acting on the object, the normal force exerted by the surface on the object, and the net force if a 10 N horizontal force is applied to the object to the right.

2)A block is pulled by two forces of 15 N and 25 N to the left, and by three forces of 10 N, 20 N, 30 N to the right. Find the magnitude and direction of the resultant force.

Question 23

The bicycle tyres are in contact with the road at X and Y. The cyclist is travelling at constant velocity on a level road. The weight of the bicycle is 180 N and the weight of the cyclist is 720 N.

(a) State the magnitude of the resultant force acting on the cyclist. Explain your answer.
(b) Define moment of a force and principle of moment

(c) Explain why the two vertical forces acting on the tyres at X and Y do not form a couple.

(d) Take moments about X to determine the size of the vertical force F acting on the tyre at Y.

(e) The cyclist leans further forward. How does this affect the force on the tyre at X? Explain your answer.

(f) A solid beam 0.5 m long is laid across a pivot to form a simple lever. The pivot is 0.1 m from the end of the beam. Calculate the heaviest load that could be lifted using a force of 500 N.

Answer 23

a. The resultant force is zero.
There is no acceleration .

b. Moment of a force=force ×perpendicular distance from point/pivot

The principle of moments states that for an object to be balanced the total clockwise moment must be equal to the total anti-clockwise moment.

c Forces are in the same direction/The forces are not opposite/The forces are not equal in magnitude
Couple acts because of two equal and opposite forces acting at a distance.
Moment is force multiplied by any distance from the body.
Turning moment is due to rotating force.

dClock wise moment =(720×0.40)+(180×0.60)
Or 396 (Nm)
Sum of clockwise moment =sum of anticlockwise moments
396 =1.3 F
F=300 n

e The force at X decreases
The force at Y increases/greater clockwise moment/FX+FY =900 N

f The greatest distance from the pivot:
0.5 − 0.1 = 0.4
Then use the values to calculate the moment:
Moment =Force × distance =500× 0.4=200Nm

Second use answer above to calculate maximum force 0.1 m from the pivot
F=M/D = 200/0.1= 2000N
The heaviest load that could be lifted by this arrangement is 2000N

Question 24

A brick of mass 3.2 kg on a sloping flat roof, at 30o to the horizontal as shown in Figure 5, slides for 2.0 m with constant acceleration down the roof in 2.0 s from the rest.

(a) Calculate:
(i) The acceleration of the brick.
(ii) The frictional force on the brick due to the roof.
(iii) A 1500 kg spaceship travels in the vacuum of space at a constant speed of 600 m/s. Ignoring any gravitational forces, what is the net force on the spaceship

Answer 24

s = ut + at^2 u=0

a = 2s/t^2 = 2 x2/2^2 = 1 ms

fd - fr where fr = ma | fd =mg sin30

mgsin30 - a = 3.2 (9.81xsin30 -1) = 12.48

In a vacuum, there is no friction due to air resistance. Newton’s first law states that an object in motion stays in motion unless acted upon by a net force. Thus, the spaceship will travel at the constant speed of 600 m/s and the net force on the spaceship must be zero as acceleration is also zero.
Since, F = ma
Therefore,
F = (1500kg)(0m/s2)
= 0 N.

Question 25

Separate into scalar and vector
Time
Energy
Speed
Power
Acceleration
Velocity
Weight

Answer 25

Answer:
Scalar:
Time
Energy
Speed
Power

Vector:
Acceleration
Velocity
Weight