(Take acceleration due to gravity $g=10 \mathrm{~ms}^{-2}$ )
As shown in figure, a $70 \mathrm{~kg}$ garden roller is pushed with a force of $\vec{F}=200 \mathrm{~N}$ at an angle of $30^{\circ}$ with horizontal. The normal reaction on the roller is
(Given $\mathrm{g=10~m~s^{-2}}$)
A force acts for 20 s on a body of mass 20 kg, starting from rest, after which the force ceases and then body describes 50 m in the next 10 s. The value of force will be:
The time taken by an object to slide down 45$^\circ$ rough inclined plane is n times as it takes to slide down a perfectly smooth 45$^\circ$ incline plane. The coefficient of kinetic friction between the object and the incline plane is :
A block of mass $m$ slides down the plane inclined at angle $30^{\circ}$ with an acceleration $\frac{g}{4}$. The value of coefficient of kinetic friction will be:
Consider a block kept on an inclined plane (incline at 45$^\circ$) as shown in the figure. If the force required to just push it up the incline is 2 times the force required to just prevent it from sliding down, the coefficient of friction between the block and inclined plane($\mu$) is equal to :
As per given figure, a weightless pulley P is attached on a double inclined frictionless surfaces. The tension in the string (massless) will be (if g = 10 m/s$^2$)
Given below are two statements :
Statement I : An elevator can go up or down with uniform speed when its weight is balanced with the tension of its cable.
Statement II : Force exerted by the floor of an elevator on the foot of a person standing on it is more than his/her weight when the elevator goes down with increasing speed.
In the light of the above statements, choose the correct answer from the options given below :
Two bodies of masses $m_{1}=5 \mathrm{~kg}$ and $m_{2}=3 \mathrm{~kg}$ are connected by a light string going over a smooth light pulley on a smooth inclined plane as shown in the figure. The system is at rest. The force exerted by the inclined plane on the body of mass $\mathrm{m}_{1}$ will be : [Take $\left.\mathrm{g}=10 \mathrm{~ms}^{-2}\right]$
A uniform metal chain of mass m and length 'L' passes over a massless and frictionless pulley. It is released from rest with a part of its length 'l' is hanging on one side and rest of its length '$\mathrm{L}-l$' is hanging on the other side of the pully. At a certain point of time, when $l=\frac{L}{x}$, the acceleration of the chain is $\frac{g}{2}$. The value of x is __________.
A block of mass M slides down on a rough inclined plane with constant velocity. The angle made by the incline plane with horizontal is $\theta$. The magnitude of the contact force will be :
A block 'A' takes 2 s to slide down a frictionless incline of 30$^\circ$ and length 'l', kept inside a lift going up with uniform velocity 'v'. If the incline is changed to 45$^\circ$, the time taken by the block, to slide down the incline, will be approximately :
A bag is gently dropped on a conveyor belt moving at a speed of $2 \mathrm{~m} / \mathrm{s}$. The coefficient of friction between the conveyor belt and bag is $0.4$. Initially, the bag slips on the belt before it stops due to friction. The distance travelled by the bag on the belt during slipping motion, is : [Take $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{-2}$ ]
Two masses $M_{1}$ and $M_{2}$ are tied together at the two ends of a light inextensible string that passes over a frictionless pulley. When the mass $M_{2}$ is twice that of $M_{1}$, the acceleration of the system is $a_{1}$. When the mass $M_{2}$ is thrice that of $M_{1}$, the acceleration of the system is $a_{2}$. The ratio $\frac{a_{1}}{a_{2}}$ will be :
Three masses $M=100 \mathrm{~kg}, \mathrm{~m}_{1}=10 \mathrm{~kg}$ and $\mathrm{m}_{2}=20 \mathrm{~kg}$ are arranged in a system as shown in figure. All the surfaces are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force $\mathrm{F}$ is applied on the system so that the mass $\mathrm{m}_{2}$ moves upward with an acceleration of $2 \mathrm{~ms}^{-2}$. The value of $\mathrm{F}$ is :
( Take $\mathrm{g}=10 \mathrm{~ms}^{-2}$ )
A monkey of mass $50 \mathrm{~kg}$ climbs on a rope which can withstand the tension (T) of $350 \mathrm{~N}$. If monkey initially climbs down with an acceleration of $4 \mathrm{~m} / \mathrm{s}^{2}$ and then climbs up with an acceleration of $5 \mathrm{~m} / \mathrm{s}^{2}$. Choose the correct option $\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)$.
For a free body diagram shown in the figure, the four forces are applied in the 'x' and 'y' directions. What additional force must be applied and at what angle with positive x-axis so that the net acceleration of body is zero?
A 2 kg block is pushed against a vertical wall by applying a horizontal force of 50 N. The coefficient of static friction between the block and the wall is 0.5. A force F is also applied on the block vertically upward (as shown in figure). The maximum value of F applied, so that the block does not move upward, will be :
(Given : g = 10 ms$-$2)
A block of mass 40 kg slides over a surface, when a mass of 4 kg is suspended through an inextensible massless string passing over frictionless pulley as shown below.
The coefficient of kinetic friction between the surface and block is 0.02. The acceleration of block is. (Given g = 10 ms$-$2.)
A block of mass M placed inside a box descends vertically with acceleration 'a'. The block exerts a force equal to one-fourth of its weight on the floor of the box. The value of 'a' will be
A block of mass 2 kg moving on a horizontal surface with speed of 4 ms$-$1 enters a rough surface ranging from x = 0.5 m to x = 1.5 m. The retarding force in this range of rough surface is related to distance by F = $-$kx where k = 12 Nm$-$1. The speed of the block as it just crosses the rough surface will be :
A system of two blocks of masses m = 2 kg and M = 8 kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocks is 0.5. The maximum horizontal force F that can be applied to the block of mass M so that the blocks move together will be :
In the arrangement shown in figure a1, a2, a3 and a4 are the accelerations of masses m1, m2, m3 and m4 respectively. Which of the following relation is true for this arrangement?
A person is standing in an elevator. In which situation, he experiences weight loss?
An object of mass 5 kg is thrown vertically upwards from the ground. The air resistance produces a constant retarding force of 10 N throughout the motion. The ratio of time of ascent to the time of descent will be equal to : [Use g = 10 ms$-$2].
A block of mass 10 kg starts sliding on a surface with an initial velocity of 9.8 ms$-$1. The coefficient of friction between the surface and block is 0.5. The distance covered by the block before coming to rest is :
[use g = 9.8 ms$-$2]
Two inclined planes are placed as shown in figure. A block is projected from the Point A of inclined plane AB along its surface with a velocity just sufficient to carry it to the top Point B at a height 10 m. After reaching the Point B the block slides down on inclined plane BC. Time it takes to reach to the point C from point A is $t(\sqrt{2}+1)$ s. The value of t is ___________.
(use $\mathrm{~g}=10 \mathrm{~m} / \mathrm{s}^{2}$ )
Explanation:
$AB = 10\sqrt 2 $ m
${v_A} = \sqrt {2 \times 10 \times 10} = 10\sqrt 2 $ m/s
${v_C} = 10\sqrt 2 $ m/s
${a_{BC}} = g\sin (30^\circ ) = 5$ m/s2
${t_{BC}} = 2\sqrt 2 \,s\,\left( {{{{v_c}} \over {{a_{BC}}}}} \right)$
${t_{AB}} = {{{v_A}} \over {5\sqrt 2 }} = 2\,s$
${t_{AB}} + {t_{BC}} = 2(\sqrt 2 + 1)$
$ \Rightarrow t = 2$
Four forces are acting at a point $\mathrm{P}$ in equilibrium as shown in figure. The ratio of force $\mathrm{F}_{1}$ to $\mathrm{F}_{2}$ is $1: x$ where $x=$ _____________.
Explanation:
${F_1} = \, + 2 \times {1 \over {\sqrt 2 }} - {1 \over {\sqrt 2 }} = {1 \over {\sqrt 2 }}$
${F_2} = 2 \times {1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }} = {3 \over {\sqrt 2 }}$
${{{F_1}} \over {{F_2}}} = {1 \over 3} = {1 \over x} \Rightarrow x = 3$
A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2 Mg force. In this situation, tension in the string is ${x \over 5}$ Mg for x = ______________. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g = acceleration due to gravity)
Explanation:
$a = {{Mg} \over {4M + M}} = {g \over 5}$ (in upward direction)
$T = M\left( {g + {g \over 5}} \right) = {{6Mg} \over 5}$
$ \Rightarrow x = 6$
A mass of 10 kg is suspended vertically by a rope of length 5 m from the roof. A force of 30 N is applied at the middle point of rope in horizontal direction. The angle made by upper half of the rope with vertical is $\theta$ = tan$-$1 (x $\times$ 10$-$1). The value of x is ____________.
(Given, g = 10 m/s2)
Explanation:
The vertical component of the tension, $T\cos\theta$, balances the weight of the mass:
$T\cos \theta = mg$
Since $ mg = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N}$:
$T\cos \theta = 100\, \text{N}$ ...... (i)
The horizontal component of the tension is given by:
$T\sin \theta = 30\, \text{N}$ ........ (ii)
Dividing equation (ii) by equation (i):
$ \frac{T\sin \theta}{T\cos \theta} = \frac{30}{100}$
Therefore:
$ \tan \theta = \frac{3}{10}$
Hence, the value of x is:
$ \therefore x = 3$
A system to 10 balls each of mass 2 kg are connected via massless and unstretchable string. The system is allowed to slip over the edge of a smooth table as shown in figure. Tension on the string between the 7th and 8th ball is __________ N when 6th ball just leaves the table.
Explanation:
At given instant
${a_{sys}} = {{6\,m \times g} \over {10\,m}} = {{6\,g} \over {10}}$
$\therefore$ ${T_{78}} = (3m) \times {a_{sys}}$
$ = (3m) \times \left( {{{6g} \over {10}}} \right)$
$ = {{3 \times 2 \times 6 \times 10} \over {10}} = 36\,N$
A block of mass 200 g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F = $\sqrt{x}$N as shown in figure.
The value of x = _____________.
Explanation:
$F \times {1 \over 2} = 0.2 \times 10 \times {{\sqrt 3 } \over 2}$
$ \Rightarrow F = 2\sqrt 3 $
$ \Rightarrow F = \sqrt {12} $ N
$\therefore$ $x = 12$
A force on an object of mass 100 g is $\left( {10\widehat i + 5\widehat j} \right)$ N. The position of that object at t = 2 s is $\left( {a\widehat i + b\widehat j} \right)$ m after starting from rest. The value of ${a \over b}$ will be ___________.
Explanation:
$\overrightarrow F = m\overrightarrow a $
$ \Rightarrow \overrightarrow a = 100\widehat i + 50\widehat j$
So, $\overrightarrow S = {1 \over 2}\overrightarrow a {t^2}$
${1 \over 2}\left( {100\widehat i + 50\widehat j} \right){2^2}$
$ = 200\widehat i + 100\widehat j$ m
so a = 200 m and b = 100 m
so ${a \over b} = 2$
The correct applied force vs distance graph will be :
If three forces ${\overrightarrow F _1},{\overrightarrow F _2}$ and ${\overrightarrow F _3}$ are represented by three sides of a triangle and ${\overrightarrow F _1} + {\overrightarrow F _2} = - {\overrightarrow F _3}$, then these three forces are concurrent forces and satisfy the condition for equilibrium.
Statement : 2
A triangle made up of three forces ${\overrightarrow F _1}$, ${\overrightarrow F _2}$ and ${\overrightarrow F _3}$ as its sides taken in the same order, satisfy the condition for translatory equilibrium.
In the light of the above statements, choose the most appropriate answer from the options given below :
$F = {F_0}\left[ {1 - {{\left( {{{t - T} \over T}} \right)}^2}} \right]$
Where F0 and T are constants. The force acts only for the time interval 2T. The velocity v of the particle after time 2T is :
(g is acceleration due to gravity)
Explanation:
On smooth incline
a = g sin30$^\circ$
by S = ut + ${1 \over 2}$at2
S = ${1 \over 2}$${g \over 2}$T2 = ${g \over 4}$T2 ........ (i)
For rough surface,
$ma = mg\sin 30^\circ - \mu mg\cos 30^\circ $
$a = g\sin 30^\circ - \mu g\cos 30^\circ $
Distance covered by the block on the rough surface in time $\alpha$T,
$s = ut + {1 \over 2}a{t^2}$
$s = 0 + {1 \over 2}(g\sin 30^\circ - \mu g\sin 30^\circ ){t^2}$
$s = {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2}$ .... (ii)
Distance covered by the block is same for both the case,
$ \Rightarrow {g \over 4}(1 - \sqrt 3 \mu ){(\alpha T)^2} = {g \over 4}{T^2}$ [from Eq. (i) and Eq. (ii)]
$ \Rightarrow 1 - \sqrt 3 \mu = {1 \over {{\alpha ^2}}} \Rightarrow \mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt 3 }}$
Comparing with $\mu = \left( {{{{\alpha ^2} - 1} \over {{\alpha ^2}}}} \right){1 \over {\sqrt x }}$
The value of the x = 3.
Explanation:
Given,
Angle of inclination, $\theta$ = 30$^\circ$
Acceleration, a = 10 ms$-$2
Acceleration due to gravity, g = 10 ms$-$2
According to the question the car and bob is as shown below,
Here, F' is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string.
In equilibrium, $\Sigma$Fx = 0 and $\Sigma$Fy = 0
$\Rightarrow$ F' cos30$^\circ$ = T sin$\alpha$
${{ma\cos 30^\circ } \over {\sin \alpha }} = T$ ...... (i)
where, m is the mass of the bob.
$F'\sin 30^\circ + mg = T\cos \alpha $
$ \Rightarrow ma\sin 30^\circ + mg = {{ma\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$ [$\because$ using Eq. (i)]
$ \Rightarrow a\sin 30^\circ + g = {{a\cos 30^\circ } \over {\sin \alpha }}(\cos \alpha )$
$10 \times {1 \over 2} + 10 = {{10 \times \sqrt 3 } \over 2}\cot \alpha $
$ \Rightarrow {1 \over 2} + 1 = {{\sqrt 3 } \over 2}\cot \alpha \Rightarrow {3 \over 2} = {{\sqrt 3 } \over 2}\cot \alpha $
or, $\cot \alpha = \sqrt 3 $
$ \Rightarrow \alpha = 30^\circ $
Explanation:
fsmax = 1a (for 1 kg block) ............ (ii)
$\mu$ $\times$ 1 $\times$ g = a
F = 15N
Explanation:
$\sqrt {{{2s} \over {{a_a}}}} = {1 \over 2}\sqrt {{{2s} \over {{a_d}}}} $ ..... (i)
${a_a} = g\sin \theta + \mu g\cos \theta $
$ = {g \over 2} + {{\sqrt 3 } \over 2}\mu g$
${a_d} = g\sin \theta - \mu g\cos \theta $
$ = {g \over 2} - {{\sqrt 3 } \over 2}\mu g$
using the above values of aa and ad and putting in equation (i) we will gate $\mu = {{\sqrt 3 } \over 5}$
Explanation:
Mbullet = 0.1 kg
V2 = 44 + 2($-$a) $\times$ 5
$ \Rightarrow $ 0 = (10)2 $-$ 2a $\times$ (0.5)
Retardation $(a) = {{1000} \over {2 \times 5}}$ = 100 m/s2
Retarding force (F) = ma = 0.1 $\times$ 100
FR = 10 N