Block 'A' of mass 100 kg is kept above another block 'B' of mass 300 kg. 'A' is tied to the wall 'C' with a horizontal string. The coefficient of friction between 'A' and 'B' is 0.35 and that between 'B' and the horizontal surface is 0.5. Find the horizontal force P, necessary to move the block 'B'.  India's Super Teachers for all govt. exams Under One Roof
Explanation: Referring to FBD of given system below Normal Reaction force on block A = N1 = mAg ---- (1) Normal Reaction force on block B = N2 = mBg + N1 = ( mAg + mBg ) ----(2) Friction force between A and B = f1 = μ1N1 = μ1mAg ---- (3) Friction force between B and horizontal surface = f2 = μ2N2 = μ2( mAg + mBg ) ---- (4) Also from FBD of block B, P = f1 + f2 = μ1mAg + μ2( mAg + mBg ) ---- (5) Given, mA = 100 kg, mB = 300 kg, μ1 = 0.35, μ2, 0.50 Putting above values in equation (5) then we get, P = 2350 N. Hence Option 3) is correct choice. India’s #1 Learning Platform Start Complete Exam Preparation
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What is the amount of work done, when a force of 10N is applied perpendicular to the direction of movement of the body?
The correct answer is 0 Nm Concept:
Calculation:
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A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is
CONCEPT:
\(⇒ W = \vec F .\vec d = Fd\ cos θ \;\;\;\;\; \ldots \left( 1 \right)\) Where F = The magnitude of the force vector, d = The magnitude of the displacement vector EXPLANATION: Given: The pull force F is applied in the antiparallel direction to the displacement of the mass on the plane.
∵ θ = 180°.⇒ cos180° = -1
So, the correct answer will be option 4. India’s #1 Learning Platform Start Complete Exam Preparation
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Two springs A and B (kA = 2kB) are stretched by applying forces of equal magnitudes at the four ends. If the energy stored in A is E, that in B is:
Concept:
\(F= -k\times x\) Where k - spring constant
\(E = \frac{1}{2}k\times x^2 \) Explanation:
FA= FB kAXA=kBXB
2kBXA=kBXB
Calculation:
\(E_B = \frac{1}{2}k_BX_B^2\) \(E_B=\frac{1}{2}\times\frac {k_A}{2}\times(2X_A)^2=2(\frac{1}{2}k_AX_A^2)\) \(E_B=2E\)
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For an applied force, if the area is less then the pressure will be-
The correct answer is More.
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When force 10N is applied on a body, it is displaced by 10 cm. What should be angle between force and displacement if work done in this process is 1J?
CONCEPT:
\(W = \overrightarrow{F}.\overrightarrow{d}=Fd~cosθ\) where W is the work done, F is the force, d is the displacement, and θ is the angle between F and d. CALCULATION: Given that W = 1J; F = 10N; d = 10cm = 0.1m; Work done is given by \(W = Fd~cosθ\) \(1 = 10\times 0.1~cosθ\) 1 = cosθ cos 0° = cosθ θ = 0°
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A bullet of 10 g mass comes out with an initial velocity of 1000 m/s and strikes the earth at a velocity of 500 m/s at the same level. What will be the work done (in water) by facing resistance to air?
The correct answer is 3750.
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A body of mass 100 gm is rotating on a circular path of radius r with constant velocity. The work done in one complete revolution will be:
CONCEPT:
If the Force F acts on a body and it gets displaced by a displacement of S, the work done in that case is given by: Work done (W) = FS Cos θ
\({\bf{Centripetal}}\;{\bf{Force}}\;\left( {\bf{F}} \right) = \frac{{m{v^2}}}{r}\)
CALCULATION: It is given that, Mass = 100 gm, radius = r
⇒ cos 90° = 0 Work done (W) = FS Cos 90° = 0 J
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Which of the following formulae is correct for work?
Option 2 is correct, i.e. W = F.S cosθ. CONCEPT:
\(W = \vec F \cdot \vec s\) Or, W = Fs cos θ
EXPLANATION: Work done (W) = F.s cosθ So option 2 is correct. India’s #1 Learning Platform Start Complete Exam Preparation
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A force \(\vec{F}=(5\hat{i}+3\hat{j})\) newton displaces a body by \((2\hat{i}-\hat{j})\) metre. The work done by the force is:
CONCEPT:
\(\Rightarrow W=Fx\times cosθ\) In vector form, \(\Rightarrow W=\overrightarrow{F}.\overrightarrow{x}\) Where W = work done, F = force, x = displacement and θ = angle between F and x CALCULATION: Given \(\vec{F}=(5\hat{i}+3\hat{j})\) and \(\overrightarrow{x}=(2\hat{i}-\hat{j})\) The work done by the force is: \(\Rightarrow W=\overrightarrow{F}.\overrightarrow{x}\) \(\Rightarrow W=(5\hat{i}+3\hat{j}).(2\hat{i}-\hat{j})\) \(\Rightarrow W=10-3\) \(\Rightarrow W=7J\) Hence, option 3 is correct. India’s #1 Learning Platform Start Complete Exam Preparation
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A girl is carrying a school bag of 4 kg mass on her back and climbs up a hill of height 500 m in 20 minutes. The work done against gravity is (g = 10 m/s2)
The correct answer is 2 × 104 J.
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When a body is moving in a circular motion, the work done by centripetal force will be:
CONCEPT:
The centripetal force acting on a body of mass 'm' revolving with radius 'r' is: \(F = {mv^2\over r}\)
Work Done (W) = Force (F) × Displacement (S) × cosθ where θ is the angle between force and displacement. EXPLANATION:
Work Done (W) = Centripetal Force (F) × Displacement (S) × cosθ W = F × S × cos90° W = 0
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The value of spring constant k depends on ________.
Option 1 : elastic properties of the material.
CONCEPT:
Spring force: In an ideal spring, the force required to stretch a string from its equilibrium position is directly proportional to the extension of the spring. This is known as Hooke's law for springs: Fs = -kx Where Fs is the spring force, x is the displacement from the equilibrium position and k is the spring constant. The spring constant is the property of the spring. EXPLANATION: Option 1: The spring constant is the characteristic property of the spring. The value of the spring constant depends on the elastic properties of the spring how the spring applies force when it is stretched. Option 2: it does not depend on the force applied. Force applied can change the extension of the spring. Option 3: Spring constant solely does not depend on the spring length. Option 4: Spring constant does not depend on the extension or compression of the spring. Extension or compression of the spring depends on the force applied. So the correct answer is option 1. India’s #1 Learning Platform Start Complete Exam Preparation
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Work done by a passenger standing on a platform holding a suitcase weighing 10 kg is:
CONCEPT:
W = F.s cosθ. Where F = force applied, s = displacement, and θ = angle between force and displacement. CALCULATION:
So displacement (s) = 0 Therefore, the work done by a person is ⇒ W = F. s cos θ = 0 J India’s #1 Learning Platform Start Complete Exam Preparation
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Work done in stretching a spring of constant k by x is \(\frac{1}{2} \ kx^2\).The work done when above spring is stretched further from x to 2x is
Option 3 : \(\frac{3}{2} \ kx^2\)
CONCEPT:
\(\Rightarrow W = \frac{1}{2} k x^{2}\) Where K = Force constant, x = Displacement of the spring
\(\Rightarrow W = \frac{K}{2} (x_{2}^{2} - x^{2}_{1})\) CALCULATION : Here x2 = 2x ,x1 = x
\(\Rightarrow W = \frac{K}{2} ((2x)^{2} - x^{2})\) \(\Rightarrow W = \frac{K}{2} (4x^{2} - x^{2})\) \(\Rightarrow W = \frac{3}{2}K x^{2}\)
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A mass M is dragged by a pulley on a horizontal plane by a force anti-parallel to its displacement. The work done in pulling the mass M is
CONCEPT:
\(⇒ W = \vec F .\vec d = Fd\ cos θ \;\;\;\;\; \ldots \left( 1 \right)\) Where F = The magnitude of the force vector, d = The magnitude of the displacement vector EXPLANATION: Given: The pull force F is applied in the antiparallel direction to the displacement of the mass on the plane.
∵ θ = 180°.⇒ cos180° = -1
So, the correct answer will be option 4. India’s #1 Learning Platform Start Complete Exam Preparation
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When we stretch a spring the work done due to the spring force is:
CONCEPT: Work:
In vector form, \(⇒ W=\overrightarrow{F}.\overrightarrow{x}\) Where W = work done, F = force, x = displacement and θ = angle between F and x EXPLANATION:
So work done is given as, ⇒ W = Fx.cosθ ⇒ W = Fx.cos180 ⇒ W = -Fx
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Choose the correct graph for total energy of the block of a spring-block system versus elongation?
Option 2 :
CONCEPT:
E = (1/2)(m × v2) where m is the mass of a body and v is the speed.
\(PE = {1\over 2}kx^2\) where k is the spring constant, x is the elongation or compression in the spring.
Total initial mechanical energy = Total final mechanical energy EXPLANATION:
\(PE = {1\over 2}kx^2\)
At mean position PE = 0 and KE = (1/2)(m × v2)
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The energy possess by the spring will:
Option 4 : Increase whether it is stretched or compressed
CONCEPT: Elastic potential energy:
\(\Rightarrow U = \frac{1}{2}k{x^2}\) Where U = elastic potential energy, k = spring constant, and x = displacement EXPLANATION:
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A force of 10 N acts on an object and the displacement is 7.5 m, in the direction of force. What is the work done in this case?
CONCEPT:
\(W = \vec F \cdot \vec s\) W = Fs cos θ Where θ = angle between force direction of motion
CALCULATION: Given - Force (F) = 10 N and displacement (s) = 7.5 m
⇒ W = F × s ⇒ W = 10 × 7.5 ⇒ W = 75 J India’s #1 Learning Platform Start Complete Exam Preparation
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Area under force-displacement graph gives:
The correct answer is option 1) i.e. Work CONCEPT:
Where F is the force acting on the object and x is the displacement caused. EXPLANATION:
\(⇒ Δ p=FΔ t\) Where Δp is the change in momentum, F is force, and Δt is the time taken.
⇒ p = mv Where m is the mass of the object and v is the velocity of the object. India’s #1 Learning Platform Start Complete Exam Preparation
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A car of mass 100 kg is moving on the horizontal ground. Find work done by the gravity when the car has moved 10 m.
CONCEPT:
⇒ Work Done (W) = Force (F) × Displacement (S) × cos θ where θ is the angle between force and displacement.
EXPLANATION:
⇒ Work Done (W) = Gravity Force (F) × Displacement (S) × cosθ ⇒ W = F × S × cos90° ⇒ W = F × S × 0 = 0
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The potential energy of the spring versus elongation graph is a/an___________.
CONCEPT:
\(PE = {1\over 2}kx^2\) where k is the spring constant, x is the elongation or compression in the spring. EXPLANATION:
\(PE = {1\over 2}kx^2\) It is the same as the graph of y = 4ax2 (which is a parabola)
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When a spring is suspended with 100 gm mass, it is stretched by 1 cm. Find the spring constant. (g = 10 m/s2)
CONCEPT:
This is known as Hooke's law for springs: Fs = -kx Where Fs is the spring force, x is the displacement from the equilibrium position and k is the spring constant.
CALCULATION: Given that spring is stretched due to 100 gm = 0.1 Kg mass. So here spring force will be weight of the mass. W = Fs = 0.1 × 10 = 1 N Spring is stretched by = 1 cm = 0.01 m Fs = W = -kx 1 = k × 0.01 k = 100 N/m So the correct answer is option 4. India’s #1 Learning Platform Start Complete Exam Preparation
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One spring has force constant 200 Nm-1, another has force constant 500 Nm-1. If they are joined in series, the force constant will be nearest to
Concept: In mechanics, two or more springs are said to be in series when they are connected end-to-end, and in parallel when they are connected side-by-side. Equivalent spring constant In Series connection: \(\frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\) In Parallel connection: \({k_{eq}} = {k_1} + {k_2}\) Calculation: Given, Force constant (k1) of one spring = 200 Nm-1 Force constant (k2) of another spring = 500 Nm-1 Since, they are connected in series \( \Rightarrow \frac{1}{{{k_{eq}}}} = \frac{1}{{{k_1}}} + \frac{1}{{{k_2}}}\) \( \Rightarrow \frac{1}{{{k_{eq}}}} = \frac{1}{{200}} + \frac{1}{{500}} \Rightarrow {k_{eq}} = \frac{{1000}}{7} = 142.8\;N{m^{ - 1}}\) ⇒ Force constant is nearest to 143 Nm-1 India’s #1 Learning Platform Start Complete Exam Preparation
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When a rotating body has a displacement of 10m due to a centripetal force of 10n the work done is
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