Laws of Motion

A block of mass 200g is kept stationary on a smooth inclined plane by applying a minimum horizontal force F $= \sqrt[]{x}$ N as shown in figure.

0 Follower 15 Views

Contributor-Level 10

For equilibrium along the incline plane is given by,

F cos 60° = mg sin 60°

$F = \frac{0.2 * 10 * \sqrt[]{3}}{2 * 1} * 2$

$= 2 \sqrt[]{3} N = \sqrt[]{12} N$

0 0

New answer posted

2 months ago

A curved in a level road has aradius 75m. The maximum speed of a car turning this curved road can be 30 m/s without skidding.; If radius of curved road is changed to 48m and the coefficient of friction between the tyres and the road remains same, then maximum allowed speed would be…….m/s.

0 Follower 3 Views

Contributor-Level 10

$f = \frac{m v^{2}}{R}$

\Rightarrow \frac{m v^{2}}{R} = μ N

$\frac{m v^{2}}{R} = μ m g$

$\Rightarrow 30 = \sqrt[]{μ * 75 * 10}$

$μ = \frac{30 * 30}{750} = 1.2$

Now, R = 48 m (new Radius of curvature)

$μ = 1.2$

$V_{m a x} = \sqrt[]{μ R g}$

$= \sqrt[]{1.2 * 48 * 10}$

$= 24 m / s e c$

0 0

New answer posted

2 months ago

A disc with a flat small bottom beaker placed on it at a distance R from its center is revolving about an axis passing through the center and perpendicular to its plane with an angular velocity $ω .$ The coefficient of static friction between the bottom of the beaker and the surface of the disc is $μ$ . The beaker will revolve with the disc if :

0 Follower 3 Views

Contributor-Level 10

F. B. D of Beaker

By NLM2

$\begin{array}{l} f = m a = m ω^{2} R \\ \Rightarrow m ω^{2} R \leq μ N \\ R \leq μ g / ω^{2} \end{array}$

0 0

New answer posted

2 months ago

A system of two blocks of masses m = 2kg and M = 8kg is placed on a smooth table as shown in figure. The coefficient of static friction between two blocked 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:

0 Follower 8 Views

Contributor-Level 10

Assume both the block is moving with common acceleration 'a' then

$a = \frac{F}{m + M} . . . . . . . (i)$

N = mg

N = 2g

$f = m a = \frac{m F}{m + M}$

For no slipping

$\Rightarrow \frac{2 F}{10} \leq g$

$F \leq 5 g$

$\begin{array}{l} F \leq 5 * 9.8 \\ F \leq 49.0 N \end{array}$

0 0

New answer posted

2 months ago

A mass of $10 k g$ is suspended by a rope of length $4 m$ , from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45^{?}$ with the vertical. Then $F$ equals: (Take $g = 10 {m s}^{- 2}$ and the rope to be massless)

0 Follower 3 Views

Contributor-Level 10

$\frac{T}{\sqrt[]{2}} = 100 \frac{T}{\sqrt[]{2}} = F$

$F = 100 N$

0 0

New answer posted

2 months ago

Two bodies of masses m₁ = 5kg and m₂ = 3kg 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 m₁ will be:

[Take g = 10 ms^-2]

0 Follower 12 Views

Contributor-Level 10

m₁ = 5kg

m₂= 3kg

As m₁ is in rest -

T = m_gg = 30

N = m₁ g cosq

T = $m_{1} g s i n θ \Rightarrow 50 s i n θ = 30$

->sinθ $= \frac{3}{5} \Rightarrow θ = 37 °$

$∴ N = 50 c o s θ = 50 c o s 37 ° = 50 * \frac{4}{5} = 40 N$

0 0

New answer posted

2 months ago

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)

0 Follower 7 Views

Contributor-Level 10

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

0 0

New answer posted

2 months ago

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]

0 Follower 3 Views

Laws of Motion physics ncert exemplar solutions class 12th chapter two

Contributor-Level 10

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

0 0

New answer posted

2 months ago

A travelling microscope is used to determine the refractive index of a glass slab. If 40 divisions are there in 1cm on main scale and 50 Vernier scale divisions are equal to 49 main scale division, then least count of the travelling microscope is __________* 10^-6 m.

0 Follower 2 Views