Class 11 Laws of Motion - Solving Problems In Mechanics

Q: Three blocks A, B and C, of masses \$4 kg, 2 kg\$ and \$1 kg\$ respectively, are in contact on a frictionless surface, as shown. If a force of \$14 N\$ is applied on the \$4 kg\$ block, then the contact force between A and B is:

\$\textbf{Given : } M_A= 4\ kg , M_B =2\ kg, M_C= 1\ kg\$\$\textbf{Step 1 : Acceleration of all the blocks }\$From figure, we see that the blocks A, B and C will move with a common acceleration a.By Newton's Second Law, Acceleration \$=\cfrac{F_{net \ External}}{M_{total}}\$ \$\Rightarrow a=\dfrac{F}{M_A+M_B+M_C}\$\$a=\dfrac{14}{4+2+1}= 2 m/s^2\$\$\textbf{Step 2 :Draw free body diagram for block A [Refer figure]}\$\$\textbf{Step 3 : Apply newton's law on block A :}\$\$\Sigma F = ma\$\$\Rightarrow 14-N_{AB}=4a_A\$Substitute the value of \$ a_A=a = 2m/s^2\$ :\$ \Rightarrow N_{AB}=14-8=6N\$Hence, the correct answer is option D.

Q: Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m?(g is the acceleration due to gravity)

All blocks are moving with constant velocity so net force on all blocks are zero.

Q: Two blocks \$A\$ and \$B\$ of masses \$3\ m \$ and \$m\$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of \$A\$ and \$B\$ immediately after the string is cut, are respectively :

Before cutting the strip\$T=mg\$after cutting the strip\$a_A=\dfrac{4mg-3mg}{3m}=\dfrac{mg}{3m}=\dfrac{g}{3}\$\$a_B=\dfrac{mg}{m}=g\$\$\dfrac{g}{3}, g\$.

Q: A string of negligible mass going over a clamped pulley of mass \$m\$ supports a block of mass \$M\$ as shown in the figure. The force on the pulley by the clamp is given by:

The horizontally acting tension , \$T=Mg\$The downward vertical force \$=(M+m)g\$So the resultant force on the pulley \$=\sqrt{[(M+m)g]^2+(Mg)^2}=g\sqrt{(M+m)^2+M^2}\$

Q: Two masses \$8\$ \$kg\$ and \$12\$ \$kg\$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

The given system of two masses and a pulley can be represented as shown in the following figure:Smaller mass, \$m_1 = 8\$ kgLarger mass,\$ m_2 = 12 \$kgTension in the string\$ = T\$Mass \$m_2\$, owing to its weight, moves downward with acceleration, and mass \$m_1\$ moves upward.Applying Newton’s second law of motion to the system of each mass:For mass \$m_1\$:The equation of motion can be written as:\$T – m_1g = ma \$… (i)For mass m2:The equation of motion can be written as:\$m_2g – T = m2a\$ … (ii)Adding equations (i) and (ii), we get:\$(m2 - m1)g = (m1 + m2)a\$ \$a = \dfrac{(m2 - m1) g}{(m1 + m2) }\$ ....(iii)\$=\dfrac{ (12 - 8)}{ (12 + 8)} × 10 = \dfrac{ 4 × 10}{ 20} = 2 ms-2\$Therefore, the acceleration of the masses is 2 m/s2.Substituting the value of a in equation (ii), we get:\$m_2g - T = m_2\dfrac{(m_2 - m_1)g }{ (m_1 + m_2)}\$\$T = (m2 - \dfrac{(m_2^2 - m1m2)}{(m1 + m2) }g\$ \$= \dfrac{2m_1m_2g }{(m_1 + m_2)}\$\$T= 2 × 12 × 8 × 10 / (12 + 8)\$\$T= 96 N\$Therefore, the tension in the string is 96 N.

Q: If two forces act in opposite direction then their resultant is \$10N\$ and if they act mutually perpendicular then their resultant is \$50N\$. Find the magnitudes of both the forces.

If the forces act as in the opposite direction then,\$F_1-F_2=10N...........1\$And if they both act as a perpendicular then \$\sqrt{F_1^2+F_2^2}=50N..............2\$From equation 1 we get \$F_1=10+F_2\$ put this value in equation 2 then we get,\$(F_2+10)^2+F_2^2=50^2\$ \$\Rightarrow F_2^2+F_2^2+100+2F_2=2500\$ \$\Rightarrow 2F_2^2+2F_2-2400=0\$ \$\Rightarrow F_2^2+40F_2-30F_2-1200=0\$ \$\Rightarrow F_2(F_2+40)-30(F_2+40)\$ \$\Rightarrow (F_2-30)(F_2+40)\$ \$\Rightarrow F_2=30N\$Put this value in equation 1 then we get,\$F_1=10+30=40N\$

Q: A body of weight \$2 kg \$ is suspended as shown in the figure The tension \$T_1\$ in the horizontal string (in kg wt) is

The tension in the above string can be divided into two components where the component opposite to weight will balance the weight while the component of this tension opposite to \${T_1}\$ will be equal to \${T_1}\$ . \$T\sin \theta = 2{\rm{kg}}\,{\rm{wt}}\$ \$T\cos \theta = {T_1}\$ The first equation gives \$T = 4{\rm{kg}}\,{\rm{wt}}\$ So \${T_1}\$ will be, \${T_1} = 2\sqrt 3 {\rm{kg}}\,{\rm{wt}}\$

Q: For the arrangement shown in the figure, the reading of spring balance is:

\$\textbf{Step 1: Acceleration of system [Refer Fig.]}\$String and spring are massless Apply Newton's second law on \$5\ kg\$ \$\Sigma F= ma\$ \$T - 5g = 5a \$ \$....(1)\$Apply Newton's second law on \$10\ kg\$ \$\Sigma F= ma\$ \$10g - T = 10a\$ \$....(2)\$Adding equation \$(1)\$ and \$(2)\$ \$a =\dfrac{5g}{15}=\dfrac{5\times 10}{15}= \dfrac{10}{3}\ m/s^2 \$ \$\textbf{Step 2: Reading of spring balance} \$Reading of spring balance will be force responsible for its elongation. Tension is responsible for the elongation of spring. From equation (2) \$10g - T = 10a \$ \$10\times 10 - T = 10 \times \dfrac{10}{3} \$ \$\Rightarrow T = \dfrac{200}{3} N \$ Reading of spring balance \$ = \dfrac{200}{3}\ N \$Hence, Option D is correct

Q: Three blocks are connected as shown in the figure, on a horizontal frictionless table and pulled to the right with a force of \$60 N\$. if \$M_1=10 kg\$, \$M_2= 20 kg\$ and \$M_3= 30 kg\$ then the value of \$T_2\$:

Let \$a\$ be the acceleration of the system.\$T_1= M_1a\$ .....(1)\$T_2 - T_1 =M_2a\$ ....(2)\$F-T_2 =M_3a\$ ......(3)Adding (1), (2) and (3) we get \$(M_1+M_2+M_3)a=F\$or \$(10+20+30)a=60 \Rightarrow a=1 m/s^2\$Now , \$T_2=(M_1+M_2)a=(10+20)(1)=30 N\$

Q: Two blocks are in contact on a frictionless table. One has a mass \$m\$ and the other \$2m\$. A force \$F\$ is applied on \$m\$.In the two cases respectively the ratio of force of contact between the two blocks will be:

If \$F\$ is the applied force then acceleration of the system is \$\dfrac{F}{{2m + m}} = \dfrac{F}{{3m}}\$ Now when we apply the force from the left, the force applied on the block \$m\$ is \$\dfrac{F}{{3m}}m = \dfrac{F}{3}\$. This will be the force in the contact. When we apply the force from the right from on the block will be \$\dfrac{F}{{3m}}2m = \dfrac{{2F}}{3}\$, this will be the force on the contact then. So the ratio is \$\dfrac{F}{3}:\dfrac{{2F}}{3} = 1:2\$ Hence, option \$B\$ is correct answer.

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Solving Problems In Mechanics

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Important Questions

Three blocks A, B and C, of masses $4 k g, 2 k g$ and $1 k g$ respectively, are in contact on a frictionless surface, as shown. If a force of $14 N$ is applied on the $4 k g$ block, then the contact force between A and B is:

Medium

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Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m?
(g is the acceleration due to gravity)

Medium

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Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and $B$ immediately after the string is cut, are respectively :

Medium

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A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by:

Medium

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Two masses $8$ $k g$ and $12$ $k g$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Hard

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If two forces act in opposite direction then their resultant is $10 N$ and if they act mutually perpendicular then their resultant is $50 N$ . Find the magnitudes of both the forces.

Medium

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A body of weight $2 k g$ is suspended as shown in the figure The tension $T_{1}$ in the horizontal string (in kg wt) is

Hard

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For the arrangement shown in the figure, the reading of spring balance is:

Medium

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Three blocks are connected as shown in the figure, on a horizontal frictionless table and pulled to the right with a force of $60 N$ . if $M_{1} = 10 k g$ , $M_{2} = 20 k g$ and $M_{3} = 30 k g$ then the value of $T_{2}$ :

Medium

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Two blocks are in contact on a frictionless table. One has a mass $m$ and the other $2 m$ . A force $F$ is applied on $m$ .
In the two cases respectively the ratio of force of contact between the two blocks will be:

Medium

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Basics of FBD

FBD and Laws of Motion - Problems L1

Three blocks A, B and C, of masses 4kg,2kg and 1kg respectively, are in contact on a frictionless surface, as shown. If a force of 14N is applied on the 4kg block, then the contact force between A and B is:

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)

Two blocks A and B of masses 3 m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively :

A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by:

Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

If two forces act in opposite direction then their resultant is 10N and if they act mutually perpendicular then their resultant is 50N. Find the magnitudes of both the forces.

A body of weight 2kg is suspended as shown in the figure The tension T1​ in the horizontal string (in kg wt) is

For the arrangement shown in the figure, the reading of spring balance is:

Three blocks are connected as shown in the figure, on a horizontal frictionless table and pulled to the right with a force of 60N. if M1​=10kg, M2​=20kg and M3​=30kg then the value of T2​:

Two blocks are in contact on a frictionless table. One has a mass m and the other 2m. A force F is applied on m. In the two cases respectively the ratio of force of contact between the two blocks will be:

Three blocks A, B and C, of masses $4 k g, 2 k g$ and $1 k g$ respectively, are in contact on a frictionless surface, as shown. If a force of $14 N$ is applied on the $4 k g$ block, then the contact force between A and B is:

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m?
(g is the acceleration due to gravity)

Two blocks $A$ and $B$ of masses $3 m$ and $m$ respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of $A$ and $B$ immediately after the string is cut, are respectively :

A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by:

Two masses $8$ $k g$ and $12$ $k g$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

If two forces act in opposite direction then their resultant is $10 N$ and if they act mutually perpendicular then their resultant is $50 N$ . Find the magnitudes of both the forces.

A body of weight $2 k g$ is suspended as shown in the figure The tension $T_{1}$ in the horizontal string (in kg wt) is

Three blocks are connected as shown in the figure, on a horizontal frictionless table and pulled to the right with a force of $60 N$ . if $M_{1} = 10 k g$ , $M_{2} = 20 k g$ and $M_{3} = 30 k g$ then the value of $T_{2}$ :

Two blocks are in contact on a frictionless table. One has a mass $m$ and the other $2 m$ . A force $F$ is applied on $m$ .
In the two cases respectively the ratio of force of contact between the two blocks will be: