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A lift of mass 1000 kg is moving with an acceleration of $1 ms^{–2}$ in upward direction Tension developed in the rope of lift is__________ N $ (g = 9.8 ms^{-2} )$
Tensile force T = m(g+a )
A rope which can withstand a maximum tension of 400 N hangs from a tree. If a monkey of mass 30 kg climbs on the rope in which of the following cases-will the rope break? (take $g =10 ms^{-2} $ and neglect the mass of rope)
Calculate F - m(g+a)
Same forces act on two bodies of different mass 2 kg and 5 kg initialy at rest. The ratio of times required to acquire same final velocity is _
$ Ft_1 = m_1 ( \nu -\nu_0) = m_1 \nu_1 $ $ Ft_2 = m_2 ( \nu -\nu_0) = m_2 \nu$ $ \therefore { t_1 \over t_2 } = { m_1 \over m_2 } $
A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of $ 4 ^\circ $ with the vertical, The acceleration of the train is ___ $ms^{-2} (g = 10 ms^{-2} ) $
$ ma = mg sin \theta $ $ \therefore a = 10 sin 4 ^\circ $ $ \therefore a = 0.7 ms^{-2} $
A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium The coefficient of friction between the chain and table is
= M/ L Weight of chain of lenght l = lg Weight of a chain of lenght ( L -l ) = (L -l ) g $ f_S = s_ R = s (L- l) g and f_S = l g $ $ \therefore l g = s. ( L - l ) g $ $ \therefore s = { l \over L - l } $
A car of mass 1000 kg travelling at 32 m/s clashes into a rear of a truck of mass 8000 kg moving in the same direction with a velocity of 4 m/s. After the collision the car bounces with a velocity of $8 ms^{–1} $. The velocity of truck after the impact is m/s
According to eqn $ m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec v_1 + m_2 \vec v_2 $ $ \therefore v_2^1 = 9 m/s $
The upper half of an inclined plane of inclination $ \theta $ is perfectly smooth while the lower half is rough A body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by_ ____
For upper half $ v_1^2 - v_o^2 = 2 ad $ $ \nu^2 - 0 = 2 ( g sin \theta )_2 $ $ \therefore v^2 = gl sin \theta $ For lower half $ v_1^2 - v_0^2 = 2 a' d $ $ 0 - gl sin \theta = 2g [ sin \theta - \mu_s cos \theta ] . { l \over 2 } $ $ \therefore 2 = 2 tan \theta $
The motion of a particle of a mass m is describe by $ y = ut + {1 \over 2 } gt^2 $ Find the force acting on the particle.
The given equation of motion is $ y = ut + \frac{1}{2}gt^2 $. This equation represents the vertical motion of a particle under the influence of gravity. The force acting on the particle is due to gravity, which is given by $ F = mg $. Hence, the correct option is F = mg.
A balloon has a mass of 10 g in air, The air escapes from the balloon at a uniform rate with a velocity of 5 cm/s and the balloon shrinks completely in 2.5 sec. calculate the average force acting on the balloon.
$ F = { dp \over dt } = { d \over dt } (mv ) = \left( {dm \over dt } \right) . \nu $ $ { dm \over dt} = { 10 \over 2.5 } = 4gram /sec and \nu = 5 cm/s $
Force= vm/t = 5×10/2.5= 20
The minimum force required to start pushing a body up a rough (cofficient of ) inclined plane is $F_1$. While the minimum force needed to prevent it from sliding down is $F_2$. If the inclined plane makes an angle $ \theta $ from the horizontal. such that $ tan \theta = 2 \mu $ than the ratio $ F_1 / F_2 $ is
$ F_1 = mg( sin\theta + cos \theta )$ $F_2 = mg(sin \theta - cos \theta) $ $ { F_1 \over F_2 } = { tan \theta + \over tan\theta - } = { 2+ \over 2- } = 3 $
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