Practice free Work Energy And Power (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
A spring with spring constant K when streched through 2cm the potential energy is U. If it is streched by 6cm. The potential energy will be......
$ Energy U \alpha x^2 ( k constant) $
If linear momentum of body is increased by 1.5%, its kinetic energy increases by……. %
$ K = { P^2 \over 2m } $ $ \therefore K \alpha p^2 $ ( m constant) $ \therefore { dk \over k } = 2 { dp \over p } $
With what velocity should a student of mass 40 kg run so that his kinetic energy becomes 160 J ?
$ k = {1 \over 2} mv^2 $ $ \therefore v = \sqrt { 2k \over m} $
A body of mass 1 kg is thrown upwords with a velocity 20 m/s. It momentarily comes to rest after a height 18m. How much energy is lost due to air friction. $(g = 10 m/s^2) $
Energy lost due to air friction $ = { 1 \over 2} mv^2 - mgh $
Two bodies of masses $m_1 and m_2$ have equal kinetic energies. If $P_1$ and $P_2$ are their respective momentum, what is ratio of $P_2 : P_1$ ?
$ P \alpha \sqrt m $ $ \therefore { p_1 \over p_2 } = { \sqrt {m_1} \over \sqrt {m_2} } $
The velocity of a body of mass 400 gm is $ ( -3 \hat i-4 \hat j) $ m/s. So its kinetic energy is ......
Find the value of $| \vec v |$ and than ,K.E = $ { 1\over 2} mv^2 $
A particle is moving under the influence of a force given by F = kx, where k is a constant and x is the distance moved. What energy (in joule) gained by the particle in moving from x = 1m to x = 3m ?
$ U = \int_1^3 kxdx $
A spring is compressed by 1 cm by a force of 4 N. Find the potential energy of the spring when it is compressed by 10 cm.
$ Spring constant K = { F \over x} , U = { 1 \over 2} kx^2 $
when 2kg mass hangs to a spring of length 50 cm, the spring stretches by 2 cm. The mass is pulled down until the length of the spring becomes 60 cm. What is the amount of elastic energy stored in the spring in this condition, if $g = 10 m/s^2 $
$ Spring constant K = { F \over x} , U = { 1 \over 2} kx^2 $
The potential energy of a projectile at its highest point is (1/2)th the value of its initial kinetic energy. Therefore its angle of projection is ......
$ Hmax = { V_o^2 Sin^2 \theta_0 \over 2g } , U = mg Hmax = { mv_0^2 sin^2 \theta_0 \over 2} = {1 \over2} Ko $
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