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Physics Laws of Motion

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Physics Laws of Motion Laws of Motion

When an elevator is moving downward with a increasing speed, the apparent weight of a body inside elevator

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alok kumar singh

Contributor-Level 10

Apparent weight = mg – ma

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Physics Laws of Motion Laws of Motion

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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alok kumar singh

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 = Ma_B . (1)

kx = Ma . (2)

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

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Physics Laws of Motion Class 11th

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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Vishal Baghel

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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Physics Laws of Motion Class 11th

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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Vishal Baghel

Contributor-Level 10

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

4a = 4g – T . (1), and

40a = T - f_k = T - $μ$ (40g) ……………… (2)

Adding equations (1) and (2), we can write

44a = 40 – 0.02 * (400) = 32

$\Rightarrow a = \frac{32}{44} = \frac{8}{11} m / s^{2}$

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Physics Laws of Motion Class 11th

A ball is released from rest point P of a smooth semi-spherical vessel as shown is figure. The ratio of the centripetal force and normal reaction on the ball at point Q is A while angular position of point Q is a with respect to point P. Which of the following graphs represent the correct relation between A and a when ball goes from Q to R?

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Raj Pandey

Contributor-Level 9

NLM at point Q

N – mg sin a = $\frac{m v^{2}}{R}$

$N = \frac{m v^{2}}{R} + m g s i n α$ - (1)

COE : between P & Q

$\frac{m v^{2}}{R} = 2 m g s i n α$ - (2)

Centripetal force = 2mg sin a

N (Normal Reaction) = 2mg sin a + mg sin a

= 3mg sin a

$\frac{C e n t r i p e t a l f o r c e}{N o r m a l R e a c t i o n} = \frac{2 m g s i n α}{3 m g s i n α} = \frac{2}{3}$

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Physics Laws of Motion Class 11th

A person is standing in an elevator. In which situation, he experiences weight loss?

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Raj Pandey

Contributor-Level 9

.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

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Physics Laws of Motion Class 11th

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 $\frac{x}{5} M g$ for x_________. Neglect mass of the string and fiction of the block (bigger mass) with ice slab.

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Raj Pandey

Contributor-Level 9

Using newton's law:

2mg – T = 4ma - (1)

T – mg = ma - (2)

From (1) & (2) :

$m g = 5 m a \Rightarrow a = \frac{g}{5} = 2 m / s^{2}$

$T = m g + \frac{m g}{5} = \frac{6 m g}{5}$

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Physics Laws of Motion Class 11th

A man of 60kg is running on the road and suddenly jumps into stationary trolly car of mass 120kg. Then, the trolly car starts moving with velocity 2ms^-¹. The velocity of the running man was _________ ms^-¹, when he jumps into the car:

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Raj Pandey

Contributor-Level 9

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒v = 6m/s

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2 weeks ago

Physics Laws of Motion Class 11th

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

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

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

0 0

New answer posted

2 weeks ago

Physics Laws of Motion Class 11th

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.)