A horizontal force F is applied on back of mass m placed on a rough inclined plane of inclination θ. The normal reaction N is

mg cos θ
mg sin θ
mg cos θ - F cos θ
mg cos θ + F sin θ

Solution

In the system shown in figure, the pulley is smooth and mass less, the string has a total mass 5g, and the two suspended blocks have masses 25 g and 15 g. The system is released from state ι = 0 and is studied up to stage ι’=0 During the process, the acceleration of block A will be

constant at ^g⁄₉
constant at ^g⁄₄
increasing by factor of 3
increasing by factor of 2

Solution

Considering the two masses and the rope a system,then

Initial net force =[25 - (15 + 5)] g = 5g

Final net force = [(25 + 5) - 15] g = 15 g

⇒(acceleration)_final= 3 (acceleration)_initial

A plate of mass M is placed on a horizontal of friction less surface (see figure), and a body of mass m is placed on this plate. The coefficient of dynamic friction between this body and the plate is μ . If a force 2μmg is applied to the body of mass m along the horizontal, the acceleration of the plate will be

^μm⁄_Mg
\(\frac{\mu m}{(M+m)}g\)
\(\frac{\mu m}{(M+m)}g\)
\(\frac{2\mu m}{(M+m)}g\)

Solution

The frictional force acting on M is µmg

∴ Acceleration = ^µmg⁄_M

A ball of mass 0.4 kg thrown up in air with velocity 301 ms^-1 reaches the highest point in 2.5 second . The air resistance encountered by the ball during upward motion is

0.88 N
8800 N
300 dyne
300 N

Solution

Let the air resistance be F. Then

mg + F = ma ⇒ F = m[a - g]

Here a = ³⁰⁄_2.5 = 12 ms^-1

A 10 kg stone is suspended with a rope of breaking strength 30 kg-wt. The minimum time in which the stone can be raised through a height 10 m starting from rest is (Take g = 10 N/kg)

0.5 s
1.0 s
\(\sqrt{2/3}\)
2 s

Solution

On a smooth plane surface (figure) two block A and B are accelerated up by applying a force 15 N on A. If mass of B is twice that of A, the force on B is

30 N
15 N
10 N
5 N

Solution

The acceleration of both the blocks = ¹⁵⁄_3x =
⁵⁄_x

∴ Force on B =⁵⁄_x × 2x = 10N

An object is resting at the bottom of two strings which are inclined at an angle of 120° with each other. Each string can withstand a tension of 20 N. The maximum weight of the object that can be supported without breaking the string is

5 N
10 N
20 N
40 N

Solution

If W is the maximum weight, then

W = 2T cos 60°

or W = T = 20N

A particle moves so that its acceleration is always twice its velocity. If its initial velocity is 0.1 ms^–1, its velocity after it has gone 0.1 m is

0.3 ms^–1
0.7 ms^–1
1.2 ms^–1
3.6 ms^–1

Solution

In the figure a smooth pulley of negligible weight is suspended by a spring balance. Weight of 1 kg f and 5 kg f are attached to the opposite ends of a string passing over the pulley and move with acceleration because of gravity, During their motion, the spring balance reads a weight of

6kg f
less then 6 kg f
more than 6 kg f
may be more or less then 6kg f

Solution

A block of mass m on a rough horizontal surface is acted upon by two forces as shown in figure. For equilibrium of block the coefficient of friction between block and surface is

\(\frac{F_{1}\, +\, F_{2}\,sin\, \theta }{mg\, +\, F_{2}\, cos\, \theta }\)
\(\frac{F_{1}\, cos\, \theta +\, F_{2} }{mg\, -\, F_{2}\, sin\, \theta }\)
\(\frac{F_{1}\, +\, F_{2} \, cos\,\theta }{mg\, +\, F_{2}\, sin\, \theta }\)
\(\frac{F_{1}\,sin\, \theta -\, F_{2}}{mg\, +\, F_{2}\, cos\, \theta }\)

Solution

Here, on resolving force F₂ and applying the concept of equilibrium

N = mg + F₂ cos θ, and f = µN

∴ f=µ [mg + F₂ cos θ ] …(i)

Also f = F₁ + F₂ sin θ …(ii)

From (i) and (ii)

µ [mg + F ₂ cos θ] = F₁ + F₂ sin θ

\(\Rightarrow \mu =\frac{F_{1}\, +\, F_{2}\, sin\, \theta }{mg\, +\, F_{2}\, cos\, \theta }\)

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