Two mass m and 2m are attached with each other by a rope passing over a friction less and mass less pulley. If the pulley is accelerated upwards with an acceleration ‘a’, what is the value of T?

\(\frac{g+a}{3}\)
\(\frac{g-a}{3}\)
\(\frac{4m(g+a)}{3}\)
\(\frac{m(g-a)}{3}\)

Solution

The equations of motion are

2 mg – T = 2ma

T– mg = ma ⇒ T =4ma & a =g/3 so T= 4mg/3

If pulley is accelerated upwards with an accleration a,then tension in string is

T=^4m⁄₃(g+a)

One end of mass less rope, which passes over a mass less and friction less pulley Pis tied to a hook C while the other end is free. Maximum tension that the rope can bear is 360 N. With what value of maximum safe acceleration (in ms^-2) can a man of 60 kg moves down wards on the rope? [Take g = 10 ms^-2]

Solution

When forces F₁, F₂, F₃ are acting on a particle of mass msuch that F₂ and F₃ are mutually perpendicular, then the particle remains stationary.If the force F₁ is now remove d then the acceleration of the particle is

F₁/m
F₂F₃/mF₁
(F₂– F₃)/m
F₂/m

Solution

The formula for force is given by F₁ = ma

Acceleration of the particle a=^F₁⁄_m,

because F₁ is equal to the vector sum of F₂ & F₃

A particle of mass 10 kg is moving in a straight line. If its displacement, x with time t is given by x = (t³– 2t – 10) m,then the force acting on it at the end of 4 seconds is

24 N
240 N
300 N
1200 N

Solution

Two trolleys of mass m and 3 mare connected by a spring.They were compressed and released at once, they move off in opposite direction and come to rest after covering a distance S₁, S₂ respectively. Assuming the coefficient of friction to be uniform, ratio of distances S₁: S₂ is :

1 : 9
1 : 3
3 : 1
9 : 1

Solution

Two blocks are connected over a mass less pulley as shown in fig. The mass of block A is 10 kg and the coefficient of kinetic friction is 0.2. Block A slides down the incline at constant speed.The mass of block B in kg is:

Solution

A 40 kg slab rests on friction less floor as shown in fig.A 10kg block rests on the top of the slab. The static coefficient of friction between the block and slab is 0.60 while the kinetic friction is 0.40. The 10 kg block is acted upon by a horizontal force of 100 N. If g = 9.8 m/s², the resulting acceleration of the slab will be:

0.98 m/s²
1.47 m/s²
1.52 m/s²
6.1 m/s²

Solution

Force on the slab (m = 40 kg) = reaction of frictional force on the upper block

∴ 40a = μk × 10 ×g or a= 0.98 m/sec²

A block of mass 5 kg resting on a horizontal surface is connected by a cord, passing over a light friction less pulley to a hanging block of mass 5 kg. The coefficient of kinetic friction between the block and the surface is 0.5. Tension in the cord is : (g = 9.8 m/sec²)

49 N
Zero
36.75 N
2.45 N

Solution

For block A, T – μN =5a and N= 5g

for block B, 5g – T = 5a

⇒T =36.75N, a = 2.45 m/sec²

In the question , the tension in the strings, when the lift is accelerating up with an acceleration 1m/sec², is

100 newton
980 newton
1080 newton
880 newton

Solution

T= m(g+a) = 100(9.8+1) = 1080N

The mass of the lift is 100 kg which is hanging on the string.The tension in the string, when the lift is moving with constant velocity, is(g = 9.8 m/sec²)

100 newton
980 newton
1000 newton
None of these

Solution

T = m (g + a) = 100 (9.8 + 0) = 980 N

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