In this page we have kinetic and potential energy worksheet . Hope you like them and do not forget to like , social share
and comment at the end of the page.
Multiple Choice Questions
Question 1
A peculiar spring has a force law
$F = -Dx^3$
which of the following is true
(a) potential energy at point x when U = 0 at x = 0 is $\frac{Dx^4}{4}$
(b) if a mass m is attached to the spring. Mass is displaced slightly, it will follow simple harmonic motion
(c) $\frac{Dx^4}{4}$ work has to be done on the spring in stretching it slowing from 0 to x
(d) none of the above
Solution
Answer is (a), (c)
$F = -Dx^3$
$dU = -F.dx$
$dU=(Dx^3)dx$
$U_x -U_0=\int_{0}^{x}{Dx^3dx}$
Or
$U_x= \frac{Dx^4}{4}$
Since
F is not equal -kx ,it will not be SHM
Now Workdone in stretching it slowing from 0 to x
$=\int_{0}^{x}{Dx^3dx}$
$=\frac{Dx^4}{4}$
Question 2
An object of mass M slides downward along a plane inclined at angle θ .
The coefficient of friction is k
Find d(U + KE)/dt
(a) kmg
^{2}cosθ (sinθ - kcosθ)
(b) kmg
^{2}sinθ (sinθ - kcosθ)
(c) kmg
^{2} (sinθ - kcosθ)
(d) none of the above
Solution
Ans. (a)
ma = mgsinθ - kmgcosθ
a = gsinθ - kgcosθ
So v = u + at
v = g (sinθ - kcosθ)t
now δ U + δ K.E = workdone by friction
d(U + E)/dt = friction force. velocity
= &mu mg cosθ * g (sinθ - kcosθ)
= kmg^{2}cosθ (sinθ - kcosθ)
Question 3
What is true for gravitational force
(a) work done between the two points does not depend on the path travelled
(b) work done in a closed loop is zero
(c) W = - ΔU
(d) it is a conservative force
Solution
Ans. (all)
Question 4
An object moves on a inclined plane in upward direction at uniform velocity. Neglect frictional force. Which of the following is true
(a) gravitational force does negative work
(b) engine does positive work
(c) normal reaction force does not do any work
(d) none of the above
Solution
Ans. (a), (b), (c)
Question 5
If F is the conservative force, U is potential energy associated with this force and dr... is displacement. Which of the following is true
(a)$\oint{\mathbf{F}.d\mathbf{r}=0}$
(b) dU.=
F.dr
(c) $\mathbf{F}= \frac{-\partial U}{\partial x}\mathbf{i}+ \frac{-\partial U}{\partial y}\mathbf{j} +\frac{-\partial U}{\partial z}\mathbf{k}$
(d) none of the above
Solution
Ans. (a),(c)
Question 6
A body(m) is moving along positive x axis at t = 0, v = v
_{0} and x = 0
It is subjected to retarded force
$F = -2x$
Find the x coordinate, where $\frac {dx}{dt}$ is 0
(a) x = v
_{0} √(m/2)
(b) x = v
_{0} √(m)
(c) x = v
_{0} √(2m)
(d) none of the above
Solution
Answer is (a)
By work energy theorem
$0-\frac{1}{2}mv_0^2=\int_{0}^{x}{(-2x)dx}$
x = v_{0} √(m/2)
Question 7
A particle moves from rest at point P
_{1} on the surface of the smooth circular cylinder of radius R
Find the angle it will leave the cylinder
(a) θ = sin
^{-1}2/3
(b) θ = cos
^{-1}1/3
(c) θ = cos
^{-1}2/3
(d) none of the above
Solution
Answer is (c)
As normal force does not do any work
$0 + mgR = \frac {1}{2}mv^2 + mgRcos \theta$
also at point of leave
$mgcos \theta = \frac {mv^2}{R}$
$v^2 = gRcos \theta$
$mgR = \frac {1}{2} mgRcos \theta + mgRcos \theta$
$1 = \frac {3}{2} cos \theta$
$cos \theta = \frac {2}{3}$
$ \theta = cos^{-1} \frac {2}{3}$
Question 8
The kinetic energy of a particle moving along a circle of radius R depend on the distance covered S as
$T = aS^2$
where a is constant
find the force acting on the particle as a function of S
(a) 2aS
^{2}/R
(b) 2aS √[1 + (S/R)
^{2}]
(c) 2aS √[1 - (S/R)
^{2}]
(d) none of the above
Solution
Answer is (b)
$T=\frac{1}{2}mv^2=aS^2$
Or
$v^2=\frac{2aS^2}{m}$
Differentiating wrt to time
$2v\frac{dv}{dt}=\frac{4aSv}{m}$
So $w_t = \frac {2aS}{m}$
normal $w_N = \frac {v^2}{R}= \frac {2aS^2}{mR}$
So net acceleration
$= \sqrt {[(\frac {2aS}{m})^2 + (\frac {2aS^2}{mR})^2]}$
$= (\frac {2aS}{m}) \sqrt {[1 + (\frac {S}{R})^2]}$
So F = 2aS √[1 + (S/R)^{2}]
Question 9
A body is thrown vertically upward with a velocity v. It rise to some height and come back. find the mean power over all cycle
(a) mgv
(b) 0
(c) mgv
^{2}
(d) none of the above
Solution
Answer is (b)
$P = -mgv$
$= -mg(v- gt)$
$= -mgv + mg^2t$
Time to reach the height=v/g
So total time of journey=2v/g
now mean power = total work/total time
$= \frac{\int_{0}^{2v/g}Pdt}{2v/g}$
= 0
Question 10
An electron and a proton are moving under the influence of mutual forces. In calculating the change in the kinetic energy of the system
during motion, one ignores the magnetic force of one on another. This is because,
(a) the two magnetic forces are equal and opposite, so they produce no net effect.
(b) the magnetic forces do no work on each particle.
(c) the magnetic forces do equal and opposite (but non-zero) work on each particle.
(d) the magenetic forces are necessarily negligible
Solution
Answer is (b)
link to this page by copying the following text
Also Read
Go back to Class 11 Main Page using below links
Class 11 Maths
Class 11 Physics
Class 11 Chemistry
Class 11 Biology