JEE PHYSICS MCQ: WORK-ENERGY-POWER

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Application +4 | -1

The potential energy of

1kg

particle free to move along the

x - axis

is given by

U\,(x) = \left[ {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right]J

. The total mechanical energy of the particle is

2J

. What is the maximum speed

\left( {in\,\,m{s^{ - 1}}} \right)

of the particle?

View answer and explanation

Correct option: A

\frac{3}{{\sqrt 2 }}m{s^{ - 1}}

Option A (

\frac{3}{{\sqrt 2 }}m{s^{ - 1}}

) is correct because it is calculated by finding the minimum potential energy and using the conservation of energy to determine the maximum kinetic energy and hence the maximum speed.

More Questions from WORK-ENERGY-POWER

Practice All Chapter MCQs →

A ball throwing cannon, lifts ball and throws with

72 \mathrm{~km} / \mathrm{hr}

Speed. If balls are raised upto

10 \mathrm{~m}

. then find Input power of cannon if it lifts 8 balls per minute and its efficiency is

80 \%

. Mass of each ball is

1 \mathrm{~kg}

. Assertion:When a particle moves in a circle with a uniform speed its acceleration is constant but the velocity changes.
Reason: Angular displacement is not an axial vector. A body of mass

m

thrown vertically upwards attains a maximum height

h

. As what height will its kinetic energy be

75 \%

of its initial value? Assertion: Centripetal force does no work.

Reason: Force and displacement are perpendicular to each other. Statement I: Both stretched and compressed springs possess potential energy.

Statement II: Work done against restoring force is stored as potential energy. A bullet of mass

m

moving at a speed

v

hits a ball of mass

M

kept at rest. A small part having mass

m^{\prime}

breaks from the ball and sticks to the bullet. The remaining ball is found to move at a speed

v_1

in the direction of the bullet. Find the velocity of the bullet after the collision.