Physics Laws of Motion

As shown in the figure, a block of mass $\sqrt[]{3} k g$ is kept on a horizontal rough surface of coefficient of friction $\frac{1}{3 \sqrt[]{3}} .$ The critical force to be applied on the vertical surface as shown at an angle 60° with horizontal such that it does not move, will be 3x. The value of x will be-------------.

[g = 10m/s² ; sin 60° = $\frac{\sqrt[]{3}}{2};$ cos60° = $\frac{1}{2}]$

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Contributor-Level 10

N = mg + F sin 60°

$f_{l i m} = F c o s 60 ° \Rightarrow μ N = F c o s 60 °$

$μ m g = F [c o s 60 ° - μ s i n 60 °]$

$F = \frac{μ m g}{c o s 60 ° - μ s i n 60 °} = \frac{\frac{1}{3 \sqrt[]{3}} * \sqrt[]{3} * 10}{\frac{1}{2} - \frac{1}{3 \sqrt[]{3}} * \frac{\sqrt[]{3}}{2}} = \frac{10 / 3}{\frac{1}{2} - \frac{1}{6}} = \frac{10 / 3}{1 / 3} = 10 N = 3 * (\frac{10}{3}) N$

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A boy pushes a box of mass 2kg with a force $\vec{F} = (20 \hat{i} + 10 \hat{j}) N$ on a frictionless surface. If the box was initially at rest, then-------------m is displacement along the x – axis after 10s.

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$a_{x} = \frac{F_{x}}{m} = \frac{20}{2} = 10 m / s^{2}$

$S_{x} = \frac{1}{2} a_{x} t^{2} = \frac{1}{2} * 10 * 1 0^{2}$

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Two masses A and B, each of mass M are fixed together by a massless spring. A force acts on the mass B as shown in figure. If the mass A starts moving away from mass B with acceleration 'a', then the acceleration of mass B will be:

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Vishal Baghel

Contributor-Level 10

Let the spring is in extension state and $a_{B} > a_{A}, {\vec{a}}_{A B} = - a_{A} \hat{i} - (- a_{B} \hat{i}) = (- a_{A} + a_{B}) \hat{i}$ $_{}_{}_{}_{}_{}_{}_{}$

Hence we can say that block moves away from block B in the frame of B

F – kx = MaB . (1)

kx = Ma. (2)

$\Rightarrow a_{B} = \frac{F}{M} - a$

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Given below are two statements : one is labelled as Assertion A and the other is labelled as Reason R.

Assertion A : Body 'P' having mass M moving with speed 'u' has head – on collision elastically with another body 'Q' having mass 'm' initially at rest. If m << M, Body 'Q' will have a maximum speed equal to '2u' after collision.

Reason R : During elastic collision, the momentum and kinetic energy are both conserved. In the light of the above statements, choose the most appropriate answer from the options given below

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Contributor-Level 10

Mass m will acquire velocity 2u. Total momentum of system will be conserved but total kinetic energy is not conserved during collision.

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Two particles having masses 4g and 16g respectively are moving with equal kinetic energies. The ratio of the magnitudes of their linear momentum is n : 2. The value of n will be……….

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Contributor-Level 10

$P = \sqrt[]{2 K m}$

$\frac{n}{2} = \sqrt[]{\frac{m_{1}}{m_{2}}} = \frac{1}{2}$

n = 1

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A bob is whirled in a horizontal plane by means of a string with an initial speed of $ω rpm$ . The tension in the string is $T$ . If speed becomes $2 ω$ while keeping the same radius, the tension in the string becomes:

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Contributor-Level 10

$T = m \hat{?} ω^{2}$

$T^{'} = m : {(2 ω)}^{2}$

$T^{'} = 4 T$

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A $10 μ F$ capacitor is connected to a $210 V, 50 Hz$ source as shown in figure. The peak current in the circuit is nearly ( $π = 3.14$ ):

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Contributor-Level 10

Capacitive Reactance $X_{C} = \frac{1}{ω C} = \frac{1}{2 π f C} = \frac{1}{2 * 3.14 * 50 * 10 * 1 0^{- 6}}$

$= \frac{1000}{3.14}$

$V_{rms} = 210 V$

$i_{rms} = \frac{V_{rms}}{X_{C}} = \frac{210}{X_{C}}$

$Peak current = \sqrt[]{2} i_{ms} = \sqrt[]{2} * \frac{210}{1000} * 3.14 = 0.932$

$? 0.93 A$

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The output $(Y)$ of the given logic gate is similar to the output of an/a

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Contributor-Level 10

$Y_{1} = \bar{A \cdot A}$

$= \bar{A}$

$Y_{2} = \bar{B + B}$

$= \bar{B}$

$Y = \bar{Y_{1} + Y_{2}}$

$= \bar{\bar{A} + \bar{B}}$

$= \bar{\bar{A}} \cdot \bar{\bar{B}}$

$= A \cdot B is similar to output of AND Gate$

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When a body slides down from rest along a smooth inclined plane making an angle of 30° with the horizontal, it takes time T. When the same body slides down from the rest along a rough inclined plane making the same angle and though the same distance, it takes time aT, where a is a constant greater than 1. The co-efficient of friction between the body and the rough plane is $\frac{1}{\sqrt[]{x}} (\frac{α^{2} - 1}{α})$ where x =___________.

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Contributor-Level 10

for smooth surface

$a = g s i n 30 ° = \frac{g}{2}$

$S_{1} = u t + \frac{1}{2} a t^{2}$

$S_{1} = \frac{1}{2} \frac{g}{2} t^{2} = \frac{g}{4} t^{2} . . . . . . . (i)$

for rough Surface

$S = \frac{1}{2} \frac{g}{2} (1 - μ \sqrt[]{3}) α^{2} t^{2} . . . . . . . . (i i)$

By (i) and (ii)

$μ = \frac{1}{\sqrt[]{3}} (\frac{α^{2} - 1}{α^{2}}) x = \sqrt[]{3}$

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Three masses M = 100kg, m₁ = 10kg and M₂ = 20kg are arranged in a system as shown in figure. All the surface are frictionless and strings are inextensible and weightless. The pulleys are also weightless and frictionless. A force F is applied on the system so that the mass m₂ moves upward with an acceleration of 2ms^-2. The value of F is:

(Take g = 10ms^-2)

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