Tension Force Questions

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Very Short Answers Questions [1 Mark Questions]

Ques. What is tension force?

Ans. Tension is a pulling force exerted by a string, along the length of the string on the object connected to the string.

Ques. What is the unit of tension?

1. Joules
2. Newton
3. Watts
4. Radian

Ans. The correct answer is b. Newton

Explanation: Tension is a type of force and its SI unit is Newton.

Ques. Tension is a type of

1. Constant force
2. Pushing force
3. Pulling force
4. None of the above

Ans. The correct answer is c. Pulling force

Explanation: Tension is the pulling force exerted by a string along its length.

Ques. In the case of tension, what happens to the work done equation W = FS?

1. W = – 1
2. W = 0
3. W = 1
4. None of the above

Ans. The correct answer is b. W = 0

Explanation: The work done by the tension force is zero.

Ques. Choose the correct answer: Tension force is a kind of

1. Contact force
2. Non-contact force

Ans. The correct answer is a. Contact force.

Explanation: Tension force is developed in a string when a force is applied to it. This can be applied by suspending a block through a string on the ceiling. Since there is physical contact between the string and the block, therefore the tension force developed is a constant force.

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Short Answers Questions [2 Marks Questions]

Ques. What is the formula of tension?

Ans. If a body of mass m is connected to a string moving upward with acceleration a, then tension in the string is given by

T = mg + ma

If a body of mass m is connected to a string moving downward with acceleration a, then tension in the string is given by

T = mg - ma

In general, the tension formula is given by

T = m(g ± a)

Ques. What are the characteristics of tension force?

Ans. The characteristics of tension force are

• It is a pulling force.
• The direction of tension force is along the length of the string.
• If the string is massless, the magnitude of tension is the same throughout the string.

Ques. What are the examples of tension force?

Ans. The applications of tension force are

• Towing a car
• Tug of war
• Pulling a rope
• Rock climbing with the rope

Ques. Is Tension force affected by gravity?

Ans. Tension force is due to the presence of electromagnetic forces which are developed in the ropes or strings when an object is connected to it. Tension force does not depend on gravity as when the rope is pulled under tension force then the objects connect in its two ends.

Also Read:

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Long Answers Questions [3 Marks Questions]

Ques. A light and inextensible string support a body of mass of 15 kg hanging from its lower end. If the upper end of the string is firmly attached to a hook on the roof, then what is the tension in the string?

Ans. Given the mass of the body, m = 15 kg

Therefore, the weight of the body acting in the downward direction, W = mg = 15 x 9.8 = 147 N

Also, an upward force acts on the body due to the tension (T) on the string.

Since the body is at rest, then the net force acting on the body is zero i.e.

T = mg = 147 N

Therefore, the tension in the string is 147 N

Ques. A monkey of mass 10 kg climbs up a light vertical string suspended from a hook with an acceleration of 2 m/s². Find the tension in the string. (take g = 10 m/s²)

Ans. Given

• The mass of the monkey, m = 10 kg
• Acceleration of the monkey, a = 2 m/s²

As the monkey moves up with acceleration, the tension (T) in the string will be equal to the apparent weight of the monkey. Therefore

T = mg + ma = m(g + a)

⇒ T = 10 (10 + 2) = 120 N

Ques. Find the tension in each string shown in the following systems.

Ans. (a) Let T be the tension in the string. The free body diagram of the block of mass m is given by

Since the block is at rest, therefore the net force acting on the block is given by

T - mg = 0

⇒ T = mg

(b) Let T₁ be the tension in the string connected between the block m₁ and m₂ and T₂ be the tension in the string connected between m₂ and the roof.

The free-body diagram of both blocks is shown below.

The net force acting on the block of mass m₁ is given by

T₁ - m₁g = 0

⇒ T₁ = m₁g

The net force acting on the block of mass m₂ is given by

T₂ - (T₁ + m₂g) = 0

⇒ T₂ = T₁ + m₂g

⇒ T₂ = m₁g + m₂g = (m₁ + m₂)g

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Very Long Answers Questions [5 Marks Questions]

Ques. A horizontal force F pulls two masses m and 2 m lying on a frictionless table and is connected by a light string. Find the tension in the string. Does the answer depend on which end the pull is applied?

Ans. From the given data, two cases can be made

1. When force is applied on mass m
2. When force is applied on mass 2m

Case 1: When force is applied on mass m

Where

• F is the applied force
• a is acceleration
• T is the tension

The free-body diagram of the block of mass m is given by

According to Newton’s second law of motion, the net force in the horizontal direction is

F - T = ma …(i)

The free-body diagram of the block of mass 2m is given by

According to Newton’s second law of motion, the net force in the horizontal direction is

T = 2ma …(ii)

Substituting T = 2ma in equation (i), we get

F - 2ma = ma

⇒ F = 3ma

⇒ a = F/3m

Substituting the above value in equation (ii), we get

T = 2m (F/3m) = 2F/3

Case 2: When force is applied on mass 2m

The free-body diagram of the block of mass m is given by

According to Newton’s second law of motion, the net force in the horizontal direction is

T = ma …(iii)

The free-body diagram of the block of mass 2m is given by

According to Newton’s second law of motion, the net force in the horizontal direction is

F - T = 2ma …(iv)

Substituting equation (iii) in equation (iv), we get

F - ma = 2 ma

⇒ F = 3ma

⇒ a = F/3m

Substituting the above value in equation (iii), we get

T = m (F/3m) = F/3

The tension in the two cases is different. Obviously, the answer depends on which mass the force is applied. The acceleration is however the same in both cases.

Ques. A body weighing 0.4 kg is whirled in a vertical circle making 2 revolutions per second. If the radius of the circle is 1.2 m, find the tension in the string when the body is

1. At the bottom of the circle
2. At the top of the circle

Ans. Given

• Mass the body, m = 0.4 kg
• The radius of the circular path, r = 1.2 m
• Frequency of the rotation, f = 2 revolutions/second

The angular frequency of the body is

ω = 2πf = 2 x 3.14 x 2 = 12.56 rad/s

The tangential velocity of the rotation, v = rω

⇒ v = 1.2 x 12.56 = 15.072 m/s

Various forces acting on the body when it is at the top and bottom positions of the circle are shown in the figure.

1. Let T_B be the tension on the string when the body is at the bottom, then according to Newton’s second law of motion, the net force acting on the body is given by

T_B - mg = ma_C

⇒ T_B = mg + ma_C

Where aC is centripetal acceleration and g is the acceleration due to gravity.

But aC = v²/r

⇒ T_B = mg + m(v²/r)

⇒ T_B = (0.4 x 9.8) + 0.4 (15.072² / 1.2) = 79.64 N

Hence, the tension in the string when the body is at the bottom of the circle is 79.64 N

2. Let T_H be the tension on the string when the body is at the top, then according to Newton’s second law of motion, the net force acting on the body is given by

T_H + mg = ma_C

⇒ T_H = ma_C - mg

⇒ T_H = m(v²/r) - mg

⇒ T_H = 0.4 (15.072² / 1.2) - (0.4 x 9.8)

⇒ T_H = 71.8 N

Hence, the tension in the string when the body is at the top of the circle is 71.8 N

Ques. A block of mass M is pulled along a horizontal frictionless surface by a rope of mass m. If the force applied to the free end of the rope is F, then what is the force transmitted to the block?

Ans. The visualization of the given data in the question is shown in figure below

Net pulling force along the direction of motion is given by

F = (M + m)a

Where a is the acceleration of the block.

∴ a = F/(M + m) …(i)

We will draw the free body diagram (FBD) of the block and the rope separately.

The force transmitted to the block is the force on the block due to tension (T) on the rope at the left end.

According to Newton’s second law of motion, the net force acting on the rope is given by

F - T = ma

⇒ T = F - ma

Substituting the value of acceleration (a) from equation (i) in the above equation, we get

T = F - mF/(M + m)

⇒ T = MF / (M + m)

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