Physics Laws of Motion

Physics Laws of Motion Laws of Motion

Contributor-Level 10

0 0

New answer posted

2 months ago

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

0 Follower 1 View

Physics Laws of Motion Laws of Motion

Contributor-Level 10

Apparent weight = mg – ma

0 0

New answer posted

2 months ago

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:

0 Follower 3 Views

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)

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

0 0

New answer posted

2 months ago

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 2 Views

Vishal Baghel

Contributor-Level 10

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

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

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

0 0

New answer posted

2 months ago

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.)

0 Follower 2 Views

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

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

0 0

New answer posted

2 months ago

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?

0 Follower 4 Views

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

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

0 0

New answer posted

2 months ago

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

0 Follower 3 Views

Contributor-Level 9

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

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

0 0

New answer posted

2 months ago

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.

0 Follower 4 Views

Contributor-Level 9

Using newton's law:

2mg – T = 4ma - (1)

T – mg = ma - (2)

From (1) & (2) :

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

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

0 0

New answer posted

2 months ago

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:

0 Follower 2 Views

Contributor-Level 9

Using conservation of momentum:

60 * v = (60 + 120) * 2

⇒v = 6m/s

0 0

New answer posted

2 months ago

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