Physics Laws of Motion

In two different experiments, an object of mass 5kg moving with a speed of 25 ms^-¹ hits two different walls and comes to rest within (i) 3 second, (ii) 5 seconds, respectively.

Choose the correct option out of the following:

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Consider a uniform cubical box of side a on a rough floor that is to be moved by applying minimum possible force $F$ at $a$ point $b$ above its centre of mass (see figure). If the coefficient of friction is $μ = 0.4$ , the maximum possible value of $100 * \frac{b}{a}$ for box not to topple before moving is

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alok kumar singh

Contributor-Level 10

For no toppling

$F (\frac{a}{2} + b) \leq m g \frac{a}{2}$

$μ \frac{a}{2} + μ b \leq \frac{a}{2}$

$0.2 a + 0.4 b \leq 0.5 a$

$0.4 b \leq 0.3 a$

$b \leq \frac{3 a}{4}$

$b \leq 0.75 a$ (in limiting case)

But is not possible as maximum value of $b$ can be equal to $0.5 a$ only.

$∴ {(100 \frac{b}{a})}_{m a x} = 50.00$

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Four forces are acting at a point P in equilibrium as shown in figure. The ratio of force F₁ to F₂ is 1: x where x = _________.

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$F_{2} = 1 c o s 45 ° + 2 c o s 45 ° = 3 c o s 45 ° = \frac{3}{\sqrt[]{2}} N$

$\begin{array}{l} F_{1} + 1 c o s 45 ° = 2 s i n 45 ° \\ \frac{F_{1}}{F_{2}} = 1 : 3 \end{array}$

x = 3

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A car is moving with speed of 150 km/h and after applying the break it will move 27m before it stops. If the same car is moving with a speed of one third the reported speed then it will stop after travelling _________m distance.

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Stopping distance = $\frac{v^{2}}{2 a} = d$ $\frac{^{}}{}$

If speed is made $\frac{1}{3} r d$ $\frac{}{}$

$d^{1} = \frac{d}{9}, d^{1} = \frac{27}{9} = 3 m$

Braking acceleration Remains same.

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A uniform chain of 6m length is placed on a table such that a part of its length is hanging over the edge of the table. The system is at rest. The co-efficient of static friction between the chain and the surface of the table is 0.5, the maximum length of the chain hanging from the table is………….m.

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Physics Laws of Motion Laws of Motion

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F.B.D. of hanging length

F.B.D. of chain lying on the table

f = T = $μ N$

$\Rightarrow λ x g = μ λ (L - x) g$

$x = 0.5 (6 - x)$

$\Rightarrow x = 3 - 0.5 x$

$1.5 x = 3$

x = 2m

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A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance $\frac{h}{2} .$ the spring constant is _________ Nm^-1. (Use g = 10 ms^-2)

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COME

$k \cdot E_{i} + P \cdot E_{i} = k \cdot E_{f} + P \cdot E_{f}$

$\Rightarrow 0 + m g (h + \frac{h}{2}) = 0 + \frac{1}{2} k {(\frac{h}{2})}^{2}$ $\Rightarrow \frac{3 h}{2} m g = \frac{1}{2} k \frac{h^{2}}{4}$

$\Rightarrow 3 m g h = \frac{k}{4} h^{2} \Rightarrow 12 \frac{m g}{h} = k \Rightarrow k = \frac{12 * 0.1 * 10}{0.1}$

k = 120Nm^-1

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A stone of mass m, tied to a string is being whirled in a vertical circle with a uniform speed. The tension in the string is

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Physics Laws of Motion Laws of Motion

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Is minimum at the highest position of the circular path.

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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 :

[used g = 9.8 ms^-2]

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a µg = 0.5 * 9.8 = 4.9 m/sec²

u = 9.8 m/sec

S =?

v = 0

v² = u² + 2as

^{$\Rightarrow 0 = 9 . 8^{2} - 2 * 4.9 s$}

^{$\Rightarrow 0 = 9.8 * 9.8 - 9.8 s$}

s = $\frac{9.8 * 9.8}{9.8} = 9.8 m$

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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

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Contributor-Level 10

According to Newton's law of motion, we can write

ma = mg – N = mg $\frac{m g}{4} = \frac{3 m g}{4}$

$\Rightarrow a = \frac{3 g}{4}$

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A block of mass 40kg slides over a surface, when a mass of 4kg 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.)

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