Forces And Matter - Physics For Class IX (Science Group) - Numericals

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Physics For Class IX (Science Group)
UNIT 5: FORCES AND MATTER
Numericals

Worked Example 1

1. A spring has spring constant k = 30 Nm–1. What load is required to produce an extension of 4 m?
Solution:
Step 1: Write down known quantities and the quantities to be found.

• k = 30 Nm^-1
• x = 4 m
• F = ?

Step 2: Write down formula and rearrange if necessary

F = kx

Step 3: Put the values in formula and calculate.

F = 30 Nm^-1 x 4 m

F = 120 N

Ans: Hence, 120 N load is required to stretch the spring by 4m.

Worked Example 2

2. Calculate the pressure at a depth of 3m in a swimming pool?

(density of water = 1000 kgm^-3).

Solution:
Step 1: Write down known quantities and quantities to be found.

• d = 3 m
• ρ = 1000 kgm^-3
• g = 10 ms^-2
• p = ?

Step 2: Write down formula and rearrange if necessary

p = dρg

Step 3: Put the values in formula and calculate

p = 3 m x 1000 kgm^-3 x 10 ms^-2

p = 30000 pa = 3.0 x 10⁴ pa

Ans: Hence 3.0 x 10⁴ pa pressure will be observed at a depth of 3 m in the swimming pool.

Worked Example 3

Worked Example 4

4. In a hydraulic lift system, what must be the surface area of a piston. If a pressure of 300 kpa is used to provide an upward force of 2000 N?
Solution:
Step 1: Write down known quantities and quantities to be found.

• P = 300 kPa = 300 × 1000 Pa = 300000 Pa = 300000 Nm^-2
• F = 2000 N
• A = ?

Step 2: Write down formula and rearrange if necessary

P = ^F/_A

A = ^F/_P

Step 3: Put the values in formula and calculate

A = 2000 N/300000 Nm^–2 = 0.00667 m²

A = 6.67 x 10^-3 m²

Ans: Hence the surface area of piston is 6.67 x 10^-3 m².

Text Book Exercise (Numericals)

Section (B) Structured Questions
Stretching of Spring
1. Some students experimented to find out how a spring stretched when loads were added to it.
Table For Load and Extension

Load (N)	Extension (cm)
0	0
2	15
4	30
6	45
8	60
10	75
12	90
14	100

a) Use these results to plot a graph. (Plot x = load, y = extension in spring).
b) Use your graph to find
i) The extension when the load is 3 N;
ii) The load which produces an extension of 40 mm.
Solution:
a) Graph between force load and extension

b) (i): The extension when the load is 3 N:

Ans: The extension in the spring is 2.3 mm when the load is 3 N.

b) (ii) The load which produces an extension of 40 mm:

Ans: The load which produces extension of 40 mm is 5.2 N.


2. The variation in extension x of the force F for a spring is shown in Fig 5.14. The point L on the graph is the elastic limit of the spring.
a) Describe the meaning of elastic limit.
b) Calculate the force in extending the spring to its elastic limit ‘L’.

Ans 2 (b): The value of force:
According to above graph,

• Extension (x) is given On Y-axis and 1 square = 1 cm. So, the elastic limit 'L' has value 7 cm.
• Force (N) is given On X-axis and 1 square = 2 N and 1 small square = 0.4 N. The point 'L' is meeting on X-axis at the value 7.6 N.

Therefore, the force in extending the spring to its elastic limit ‘L’ is 7.6 N

Hooke's Law
4. Calculate the spring constant for a spring which extends by a distance of 3.5cm when a load of 14 N is hung from its end.
Ans: Solution:
Data:

• Distance x = 3.5 cm = ^3.5/₁₀₀ =0.035 m
• Force = F = 14 N
• Spring constant = k = ?

Working Formula:

• F = kx

Calculation:

14 N = k (0.035 m)

0.035 k = 14

k = ¹⁴/_0.035
k = 0.0025 Nm^-1 = 2.5 x 10^-3 Nm^-1

Ans: Therefore, the spring constant for a spring is 2.5 x 10^-3 Nm^-1.

5. Table 5.4 shows the results of an experiment to stretch a spring.

Table 5.4 For load and extension in spring

Load (N)	Extension (cm)
0.0	0.0
2.0	80.0
4.0	83.0
6.0	86.0
8.0	89.0
10.0	92.0
12.0	93.0

a) Use the result to plot an extension against load graph.
b) On the graph mark the limit of proportionality and state the value of the load at this point.
c) Calculate the spring constant k.
d) Show maximum force at which hooke's law is applicable.
Ans 5 (a): Graph Between Extension (mm) and Load (N)

Ans 5 (b): Graph Showing The Limit Of Proportionality:

• The limit of proportionality is 92 mm.
• The value of the load at this point is 10 N

Ans 5 (c):The spring constant k:
Solution:
Data:

• Force = F = 10 N
• Distance = x = 92 mm
• Spring constant = k =?

Working Formula:

F = kx

Calculation:

10 = k (92)

92 k = 10

k = ¹⁰ /₉₂

k = 0.1086 Nm^-1 = 1.086 x 10^-1 Nm^-1

Ans: Therefore, the spring constant k is 1.086 x 10^-1 Nm^-1

Ans 5 (d):The maximum force at which Hooke's law is applicable is 10 N .

8. A boy is pressing a thumbtack into a piece of wood with a force of 20 N. The surface area of head of thumbtack is 1 cm² and the cross-section area of the tip of the thumbtack is 0.01 cm². Calculate:
a) The pressure exerted by boy's thumb on the head of thumbtack.
b) The pressure of the tip of the thumbtack on the wood.
c) What conclusion can be drawn from answers of part (a) and (b)?
Ans: Solution:

Ans 8 (c):
The pressure exerted on a surface by an object increases as the weight of the object increases or the surface area of contact decreases as the weight of the object decreases or the surfsce area of the contact increases.

9. The Fig 5.15 shows a basic hydraulic system that has small and large pistons of cross section area of 0.005 m² and 0.1 m² respectively. A force of 20 N is applied to small piston. Calculate:
a) The pressure transmitted into hydraulic fluid.
b) The force at large piston.
c) Discuss the distance travelled by small and large pistons.

Ans 9 (a): Solution:
Data:

• Area of small piston: A₁ = 0.005 m²
• Area of large piston: A₂ = 0.1 m²
• Force on small piston = F₁ = 20 N
• Pressure = P = ?

Ans: The pressure transmitted into hydraulic fluid is 4000 Nm^-2 or 4.0 x 10³ Nm^-2.

Ans 9 (b): Solution:
Data:

• Area of small piston: A₁ = 0.005 m²
• Area of large piston: A₂ = 0.1 m²
• Force on small piston = F₁ = 20 N
• Force on large piston = F₂ = ?

Ans: The force at large piston is 400 N.

Ans 9 (c): We know that according to Pascal's law:

P₁ = P₂

This allow the lifting of a heavy load with a small force but there can be no multiplication of work, so in an ideal case with no frictional loss.

W_input = W_output

W₁ = W₂

We know that, Work done = Force x distance = F x d

W₁ = W₂

Therefore, F₁d₁ = F₂d₂

We have to pay for the multiplied output force by exerting the smaller input force through a large distance. Distance covered by the large piston is smaller as compared to the small piston.

Self Assessment Questions (Numericals)

Q.1: An elastic spring is 70 cm long. When it is stretched by hanging some load its length increases to 100 cm. Calculate its extension?
Solution:
Data:

• Original length of the spring = 70 cm
• Length of the stretched spring = 100 cm
• Extension = ?

Formula:

Length of stretched spring = Original length + Extension

Calculation:

100 = 70 + Extension

Extension = 100 - 70 = 30 cm

Ans:Therefore, its extension is 30 cm.

Q.2: Following table shows the results of an activity to stretch an elastic spring. Complete the table and draw a graph to represent this data.

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	00	0.0
1.0	30	1.0
2.0	32	2.0
3.0	34	3.0
8.0	36	4.0
4.0	38	5.0
5.0	40	6.0
6.0	41.5	8.0
7.0	42	10
8.0	43	10

Solution:
Given Data:

• Load (N) = 0.0, 1.0, 2.0, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0
• Length (cm) = 30, 32, 34, 36, 38, 40, 41.5, 42, 43

Required Data:

• Extension (cm) = ?
• Graph = ?

Table for load and extension

Load (N)	Length (cm)	Extension (cm)
0.0	30	0.0
1.0>	32	2.0
2.0	34	4.0
3.0	36	6.0
4.0	38	8.0
5.0>	40	10
6.0	41.5	11.5
7.0	42	12
8.0	43	13

GRAPH:

Result: The graph shows that extension depends on the load, the extension increases in equal steps as the load increases.

Q.3: How much force is needed to pull a spring to a distance of 30 cm, the spring constant is 15 Nm^-1?
Solution:
Data:

• x = 30 cm = ³⁰/₁₀₀  = 0.3 m
• k = 15 Nm^-1
• F = ?

Formula:

F = kx

Calculation:

F = 15 x 0.3 = 4.5 N

Ans:Therefore, 4.5 N force is needed to pull a spring to a distance of 30 cm.

Q.8: A wooden block of dimensions 0.5 m × 0.6 m × 1.0 m kept on the ground has a mass of 200 kg. Calculate the maximum pressure acting on the ground.
Ans: Solution:
Data:

• Dimensions of the wooden block = 0.5 m × 0.6 m × 1.0 m
• Area = A = l x b x h = 0.5 m × 0.6 m × 1.0 m = 0.3 m
• mass = 200 kg
• Acceleration due to gravity = 9.8 m/s²
• Force = weight = mg = (200) x (9.8) = 1960 N

Required Data:

• Maximum Pressure = ?

Ans:Therefore, the maximum pressure acting on the ground is 6533.33 pa

Q.9: If the density of sea water is 1150 kgm^-3, calculate the pressure on a body of 50 m below the surface of sea?
Solution:
Data:

• Density of sea water = ρ = 1150 kgm^-3
• Depth = 50 m
• Acceleration due to gravity = 9.8 m/s²

Required Data:

• Pressure = P = ?

Working Formula:

P = dρg

Calculation:

P = (1150) x (50) x (9.8)
P = 563500 pa

Ans: Therefore, the pressure on a body of 50 m below the surface of sea is 563500 pa.

Q.11: In a hydraulic press, a force of 100 N is applied on the pump of cross-sectional area 0.01 m². Find the force that compresses a cotton bale placed on larger piston of cross-sectional area 1 m².
Solution:
Data:

• F₁ = 100 N
• A₁ = 0.01 m²
• A₂ = 1 m²

Required Data:

• F₂ =?

Ans: Therefore, the force that compresses a cotton bale placed on larger piston of cross-sectional area 1 m² is 1 x 10⁴ N.