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Physics Laws of Motion

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3 weeks ago

Physics Laws of Motion Class 11th

The coefficient of static friction between a wooden block of mass 0.5kg and a vertical rough wall is 0.2. The magnitude of horizontal force that should be applied on the block to keep it adhere to the wall be__________ N.

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alok kumar singh

Contributor-Level 10

f = W = 0.5 * 10 = 5 N

N = F

For block not to slide,

$f \leq μ N$

$\begin{array}{l} \Rightarrow 5 \leq 0.2 F \\ \Rightarrow F \geq 25 N \end{array}$

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3 weeks ago

Physics Laws of Motion Class 11th

A force on an object of mass 100g is $(10 \hat{i} + 5 \hat{j}) N .$ The position of that object at t = 2s is $(a \hat{i} + b \hat{j}) m$ after starting from rest. The value of $\frac{a}{b}$ $\frac{}{}$ will be………

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Raj Pandey

Contributor-Level 9

$\vec{F} = 10 \hat{i} + 5 \hat{j}$

$\vec{a} = \frac{10}{m} \hat{i} + \frac{5}{m} \hat{j}$

$= \frac{10}{0.1 k g} \hat{i} + \frac{5}{0.1} \hat{j}$

$\vec{a} = 100 \hat{i} + 50 \hat{j}$

$a_{x} = 100, a_{y} = 50$

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

$= 0 * 2 + \frac{1}{2} * 100 * 2 * 2$

Sx = 200 m

$\frac{a}{b} = \frac{200}{100} = 2$

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3 weeks ago

Physics Laws of Motion Class 11th

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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Raj Pandey

Contributor-Level 9

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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New answer posted

3 weeks ago

Physics Laws of Motion Class 11th

If force $\vec{F} = 3 \hat{i} + 4 \hat{j} - 2 \hat{k}$ acts on a particle having position vector $2 \hat{i} + \hat{j} + 2 \hat{k}$ then, the torque about the origin will be:

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Raj Pandey

Contributor-Level 9

$\vec{r} = 2 \hat{i} + \hat{j} + 2 \hat{k}$

$\vec{τ} = \vec{r} * \vec{F}$

$= (2 \hat{i} + \hat{j} + 2 \hat{k}) * (3 \hat{i} + 4 \hat{j} - 2 \hat{k}) = | \begin{matrix} \hat{i} & - \hat{j} & \hat{k} \\ 2 & 1 & 2 \\ 3 & 4 & - 2 \end{matrix} |$

$\vec{τ} = - 10 \hat{i} + 10 \hat{j} + 5 \hat{k}$

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3 weeks ago

Physics Laws of Motion Class 11th

Consider two spring balances hooked as shown in the figure. We pull them in opposite directions with equal force. If the reading shown by A is 5 N, reading shown by B will be ______ N.

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Vishal Baghel

Contributor-Level 10

Both will show same tension
∴ Reading of is = 5 N.

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3 weeks ago

Physics Laws of Motion Class 11th

Shown in figure is a block at rest having an inclined smooth groove in vertical plane. The horizontal surface is smooth. A ball of same mass as that of block at rest is released from top end. The time when ball will leave groove is √(nL/(2g√3)). Find n.

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Vishal Baghel

Contributor-Level 10

-mv cos 60? + 2mu = 0 => v = 4u
½m [v² + u² + 2uv cos 120? ] + ½mu² = mgx sin 60?
=> v² = (8/7)√3gx => ar = (4√3/7)g
∴ t = √ (2L * 7)/ (4√3g)

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3 weeks ago

Physics Laws of Motion Class 11th

A solid circular disc (m, R) translates with V? and rotates with ω? = 2V₀/R at t = 0. Because of friction, pure rolling starts at t = t₀. The coefficient of friction is μ. Find t₀.

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Vishal Baghel

Contributor-Level 10

a = fr/m
∴ V = V? + at? = V? + (fr/m)t?
and α = 2Rfr/mR² = 2fr/mR
∴ ω = ω? - (2fr/mR)t?
∴ V = ωR ∴ v? + (fr/m)t? = ω? R - (2fr/m)t?
= (3fr/m)t? = V? = (fr/m)et?
∴ V = V? + V? /3 = 4V? /3
t? = mV? / (3μmg)
= V? /3μg

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3 weeks ago

Physics Laws of Motion Laws of Motion

A car starts from rest and accelerates at 5 m/s². At t = 4 s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at
t = 6 s ? (Take g = 10 m/s² )

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alok kumar singh

Contributor-Level 10

velocity of car at t = 4sec is
v = u + at
v = 0 + 5 (4)
= 20 m/s
At t = 6sec
acceleration is due to gravity ∴ a = g = 10 m/s
v? = 20 m/s (due to car)
v? = u + at
= 0 + g (2) (downward)
= 20 m/s (downward)
v = √ (v? ² + v? ²)
= √ (20² + 20²)
= 20√2 m/s

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New answer posted

3 weeks ago

Physics Laws of Motion Class 11th

Calculate the maximum acceleration of a moving car so that a body lying on the floor of the car remains stationary. The coefficient of static friction between the body and the floor is $0.15 (g = 10 {m s}^{- 2})$ .

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Raj Pandey

Contributor-Level 9

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New answer posted

4 weeks ago

Physics Laws of Motion Class 11th

A football player is moving southward and suddenly turns eastward with the same speed to avoid an opponent. The force that acts on the player while turning is:

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Raj Pandey

Contributor-Level 9

Initial velocity $= - v \hat{j}$

Final velocity $= - v \hat{j}$

Change in velocity $= v \hat{i} - (- v \hat{j})$

$= v (\hat{i} + \hat{j})$ Momentum gain is along $\hat{i} + \hat{j}$

$\Rightarrow$ Force experienced is along $\hat{i} + \hat{j}$

$\Rightarrow$ Force experienced is in North-East direction.

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