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The kinetic energy of a body varies directly as square of time elapsed. The force acting varies directly as

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Contributor-Level 10

$K . E . \propto t^{2}$

$\begin{array}{l} \Rightarrow \frac{1}{2} m v^{2} \propto t^{2} \\ \Rightarrow v \propto t \\ \Rightarrow \frac{d v}{d t} = c o n s t a n t \\ \Rightarrow F = m a = \frac{m d v}{d t} = c o n s t a n t \end{array}$

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2 weeks 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

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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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2 weeks 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.)

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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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2 weeks 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?

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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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2 weeks ago

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

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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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2 weeks 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.

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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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New answer posted

2 weeks 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

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alok kumar singh

Contributor-Level 10

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

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

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

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New answer posted

2 weeks 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.)

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alok kumar singh

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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New answer posted

3 weeks ago

A 20 kg block B is suspended from a cord attached to a 40 kg cart A. Find the ratio of the acceleration of block in cases (i) and (ii) shown in the figure immediately after the system is released from rest. (Neglect friction)

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Contributor-Level 9

Case I:

$T - N = 40 a$

and $20 g - T = 20 a$

Also $N = 20 a$

After simplifying, we get

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

Acceleration of block $B, = \sqrt[]{2} a = \frac{g}{2 \sqrt[]{2}} .$

Case II:

$T = 40 a$

and $20 g - T = 20 a$

After simplifying above equation, we get

$a = g / 3$

Ratio $= \frac{g / 2 \sqrt[]{2}}{g / 3} = \frac{3}{2 \sqrt[]{2}} .$

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3 weeks ago

The new white belt of a long horizontal conveyor is moving with a constant speed v = 3.0m/s. A small block of carbon is placed on the belt with zero initial velocity relative to the ground. The block will slip a bit before moving with the belt, leaving a black mark on the belt (figure). How long is that mark (in m) if the coefficient of kinetic friction between the carbon block and the belt is 0.20 and the coefficient of static friction is 0.30?

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