50 questions · timed · auto-graded
Explanation:
Kinetic energy is possessed by a body by virtue of its state of motion. So, a body at rest cannot possess kinetic energy. A body at rest will possess potential energy. Thermal and magnetic energies are irrespective of state of rest or of motion of a body.
Explanation:
Moving vehicle on the road possesses kinetic energy.
A running athlete possesses kinetic energy.
Stone on the road possesses kinetic energy.
A stretched rubber band possesses potential energy.
Explanation:
When a ball is projected upwards it's height increases.
As height increases, v velocity decreases (Kinetic Energy Decreases) so Potential energy increases.
Potential Energy = mgh
Explanation:
We know that, P.E = mgh
So, It is directly proportional to height and mass.
Since both the weights are at the same height, so the weight with a larger mass will have more potential energy.
Since 10kg object has a larger mass than a 5 medical history
So, the potential energy of a 10kg mass will be greater.
Explanation:
When the force on a particle is always perpendicular to its velocity, the work done by the force on the particle is zero, as the angle between the force and velocity is 90°. So, kinetic energy of the particle will remain constant. The force acting perpendicular to the velocity of the particle provides centripetal acceleration that causes the particle to move in a circular path.
Solution:
As power, P = force × veclocity
$\text{P}=\big[\text{MLT}^{-2}\big]\big[\text{LT}^{-1}\big]=\big[\text{ML}^2\text{T}^{-3}\big]$
As, $\text{P}=\big[\text{ML}^2\text{T}^{-3}\big]$
= Constant
$\therefore\text{ L}^2\text{T}^3=\text{Constant}$
Or, $\frac{\text{L}^2}{\text{T}^3}=\text{Constant}$
$\therefore\text{ L}^2\propto\text{T}^3$
Or, $\text{L}\propto \text{T}^{\frac{3}{2}}$
Hence, right choice is (iii) $\text{t}^{\frac{3}{2}}$
Explanation:
Taking + x direction to be positive, and assuming ball was travelling in + x direction initially.
Pi = Mv
After collision ball will move in - x direction
Pf = − Mv
Change in momentum:
ΔP = Pi − Pf
ΔP = Mv + Mv = 2Mv
Explanation:
When you compress a spring, it possess potential energy. The force of compression is proportional to the compression, according to Hooke's Law. Releasing the spring turns the potential energy into kinetic energy. The spring can be then used to propel some object.
Explanation:
Suppose that one end of an extensible string is attached to a mass m, while the other end is fixed. The mass moves with a velocity v in a vertical circle of radius R. At some instant, the string makes an angle $\theta$ with the vertical as shown in the figure.
For a complete circle, the minimum velocity at L must be $\nu_\text{L}=\sqrt{5\text{gl}}.$
Applying the law of conservation of energy, we have:
Total energy at M = total energy at L
i.e., $\frac{1}{2}\text{m}\nu_{\text{M}^2}+\text{mgl}=\frac{1}{2}\text{m}\nu_{\text{L}^2}$
$\Rightarrow\frac{1}{2}\text{m}\nu_{\text{M}^2}=\frac{1}{2}\text{m}\nu_{\text{L}^2}-\text{mgl}$
Using $\nu_\text{L}\geq\sqrt{5\text{gl}},$ we have:
$\frac{1}{2}\text{m}\nu_{\text{M}^2}\geq\frac{1}{2}\text{m}(5\text{gl})-\text{mgl}$
$\therefore\ \nu_\text{M}=\sqrt{3\text{gl}}$
Conservative forces are those forces in which work done (i) in a closed path is zero and (ii) is independent of path.
Conservative forces: Gravitational and Electrostatic force.
Non-conservative forces: Frictional force and air resistance.
Explanation:
The energy exerted or work done by our muscles on the rubber band is consumed in changing its shape. It gets stored in the stretched rubber band as its potential energy. It is this stored energy that is used by the rubber band to move to its original state, shape, and size. When the rubber band is released, this stored potential energy gets converted into kinetic energy.
If a pebble is placed in contact with the stretched rubber band, this kinetic energy is transferred to the pebble. This kinetic energy of the pebble is enough to do some destructive work, like breaking a glass window, injuring someone, etc.
Explanation:
As the distance keeps on decreasing and there will be a deceleration in the blocks.
In further results, collision increases frequently many times and slowly come in contact.
Explanation:
When a stone is placed at the roof it is at a certain height that is given by In the = m × g × h
As we have a certain value for h there would be some value for potential energy.
As the stone is at rest at the roof it will not have any kinetic energy as its velocity is zero, $\text{K}=\frac{1}{2}(\text{m}\times\text{in}^2)=0$
Explanation:
Here, a ball of mass m moving with velocity in collides elastically with wall hence, momentum is.
pi = min
The ball rebounds from wall hence, final momentum is.
pf = −min
Change in momentum is.
Δp = pi − pf
Δp = mv − (−min) = 2min
Explanation:
The potential energy of a two particle system depends only on the separation between the particles.
Explanation:
By definition, $\text{p}\frac{\text{dW}}{\text{dt}}$
$\because$ Work done = Kinetic energy
$\Rightarrow\text{p}=\frac{\text{dW}}{\text{dt}}$
$=\frac{\text{d(KE)}}{\text{dt}}=\text{constant}$
For an elastic spring, the work done is equal to the negative of the change in its potential energy.
When the spring was initially compressed or stretched by a distance x, its potential energy is given by,
$(\text{P.E.})_\text{i}=\frac{1}{2}\text{kx}^2$
When it finally comes to its natural length, its potential energy is given by,
$(\text{P.E.})_\text{f}=0$
$\therefore$ Work done $=-[(\text{P.E.})_\text{f}-(\text{P.E.})_\text{i}]=-\Big[0-\frac{1}{2}\text{kx}^2\Big]$
$=\frac{1}{2}\text{kx}^2$
Explanation:
No, the earth does not perform any work on the moon.
Work done(W) is defined as the scalar product of force(F) and displacement(s).
So, W = F × s = FsCosI where is the angle between force and displacement vector.
The radial force exerted on the moon by earth i.e attractive force due to gravity acts in direction perpendicular to which the moon suffers the displacement during rotation.
So, I = 90 hence CosI = 0
So, W = |F|| s| × 0 = 0
Explanation:
We know that Potential energy is the energy possessed by a body by the virtue of its position.
$\therefore$ An overhead tank having some water possesses Potential energy, as it is at a height.
Explanation:
The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. So, the hanging coconut has potential energy due to its location (height).
Explanation:
Explanation:
When external forces acting on a system have zero resultant, the centre of mass may move with a constant velocity i.e. it must not accelerate.
The path taken by the suitcase.
The time taken by you in doing so.
Your weight.
Work done by us on the suitcase is equal to the change in potential energy of the suitcase.
i.e., W = mgh
Here, mg is the weight of the suitcase and h is height of the table.
Hence, work done by the conservative (gravitational) force does not depend on the path.
Explanation:
The potential energy possessed by the object is the energy present in it by virtue of its position or configuration. Water stored in a dam, when allowed to flow, has kinetic energy which was earlier stored as potential energy.
Explanation:
Kinetic energy is due to the motion of the object and potential energy is due to relative elevation.
So as the plane takes off height increases and also its speed increase, so both kinetic and potential energy increases.
Kinetic energy$=\frac{1}{2}\text{min}^2$
Potential energy = mgH
Explanation:
Consider that the stone is projected with initial speed v.
As the stone is falls under the gravitational force, which is a conservative force, the total energy of the stone remains the same at every point during its motion.
From the conservation of energy, we have:
Initial energy of the stone = final energy of the stone
i.e., (K.E.)i + (P.E.)i = (K.E.)f + (P.E.)f
$=\frac{1}{2}\text{mv}_\text{r}^2+\text{mgh}=\frac{1}{2}\text{m}(\text{v}_\text{max})^2$
$\Rightarrow\text{v}_\text{max}=\sqrt{\text{v}^2+2\text{gh}}$
From the above expression, we can say that the maximum speed with which the stone hits the ground depends on the speed of projection and greater than it.
Explanation:
Conservation of Momentum principle, Initial momentum is Mi = 5m1
Final momentum is Mf = 4(m1 + m2)
By the above stated principle Mi = Mf = 5m1 = 4(m1 + m2)
⟹ m1 = 4m2
$\therefore$ m1 : m2= 4 : 1
Explanation:
When work is done by an external forces on a system, the total energy of the system will change.
Explanation:
mass of girl M = 50kg
h = 1. 2m
A girl is jumping vertically upward, when it will reach at maximum Hight its velocity will become zero ie inf = 0
K. And $=\frac{1}{2}\text{min}^2=\frac{1}{2}\text{m}\times0=0$
Explanation:
Potential energy is stored in a wound spring. Potential energy is a type of mechanical energy. Hence, energy present in a wound spring is mechanical energy.
Explanation:
$\text{power}=\frac{\text{Work}}{\text{Time}}=\frac{\text{mgh}}{\text{t}}$
Here, m = 100kg, h = 10m
and t = 1 min = 60s
$\therefore\text{p}=\frac{100\times9.8\times10}{60}=163.3\text{W}$
Explanation:
The sphere is rolling on a smooth surface i.e., no energy is lost to friction and the kinetic energy of sphere A is transferred to the sphere B without any loss as the collision is head-on and elastic.
Thus, the kinetic energy of sphere A and sphere B are equal their ratio will be 1 : 1
Explanation:
When an arrow is drawn back by a bow, the work done by us in stretching the bowstring gets stored at potential energy in the bow.
This potential energy of bow is transformed into kinetic energy when the bowstring is released and this gives kinetic energy to the arrow.
Explanation:
Constant after some depth. This constant velocity is called terminal velocity, hence KE become constant beyond this depth, which is best represented by (b).
Explanation:
The energy which a body possesses by virtue of being in motion is called as kinetic energy.
If we want to stop a moving body then we must do some work on the body to stop it which means that a moving body possesses energy because of motion.
If an object of mass m moving with the velocity v, The kinetic energy is defined as,
$\text{K.E}=\frac{1}{2}\text{mv}^2$
Explanation:
Mass of the boy, m = 40kg
Number of steps = 50
Height of each step = 10cm
Force on the boy due to gravity, F = mg = 40 × 9. 8 N = 392N
While climbing up the steps, the boy does work against gravity.
Displacement in the vertical direction, s = (50 × 10)cm = 500cm = 5m
Displacement is in the direction of force applied by the boy against gravity.
So, work done, In = F × s = 392 × 5J = 1960J
If the collision is elastic, which of the following is a possible result after collision?
Explanation:
Key concept: In a collision if the motion of colliding particles before and after the collision is along the same line, the collision is said to be head on or one dimensional.
When two bodies of equal masses collides elastically, their velocities are interchanged. Kinetic energy and linear momentum remains conserved Total kinetic energy of the system before collision
$=\frac{1}{2}\text{mv}^2+0=\frac{1}{2}\text{mv}^2$
In (a), kinetic energy of the system after collision,
$\text{K}_1=\frac{1}{2}(2\text{m})\Big(\frac{\text{v}}{2}\Big)^2=\frac{1}{4}\text{mv}^2$
Hence this option is incorrect.
In (b), kinetic energy of the system after collision,
$\text{K}_2=\frac{1}{2}(\text{m})(\text{v})^2=\frac{1}{2}\text{mv}^2$
Hence this option will be correct.
In (c), kinetic energy of the system after collision,
$\text{K}_3=\frac{1}{2}(3\text{m})\Big(\frac{\text{v}}{3}\Big)^2=\frac{1}{6}\text{mv}^2$
Hence this option is incorrect.
In (d), kinetic energy of the system after collision,
$\text{K}_4=\frac{1}{2}\text{mv}^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{2}\Big)^2+\frac{1}{2}\text{m}\Big(\frac{\text{v}}{3}\Big)^2=\frac{49}{72}\text{mv}^2$
We see that kinetic energy is conserves only in (b).
Explanation:
The kinetic energy of a body is the energy by virtue of its motion. This is the definition of kinetic energy. If it has motion then it will have velocity, hence kinetic energy exists.
$\text{K.E}=\frac{1}{2}\text{mv}^2$
Explanation:
We know that force $\text{F}=\frac{\Delta\text{p}}{\Delta\text{t}}$ and $\frac{\Delta\text{p}}{\Delta\text{t}}=\frac{\text{m}[(-\text{v})-\text{u}]}{\text{t}}=\frac{-2\text{mv}}{\text{t}}$
And the magnitude of force will be, $\text{F}=\frac{2\text{mv}}{\text{t}}$
According to the problem, $\text{m}=150\text{g}=\frac{150}{1000}\text{kg}=\frac{3}{20}\text{kg}$
$\Delta\text{t}=\text{time of contact}=0.001\text{s}$
$\text{u}=126\text{km/ h}=\frac{126\times1000}{60\times60}\text{m/ s}=35\text{}\text{m s}$
$\text{v}=-126\text{km/ h}=-35\text{m/ s}$
Change in momentum of the ball
$\Delta\text{P}=\text{m(v}-\text{u})$
$=\frac{3}{20}(-70)=-\frac{21}{2}$
We know that force $\text{F}=\frac{\Delta\text{P}}{\Delta\text{t}}$
$=\frac{\frac{-21}{2}}{0.001}\text{N}=-1.05\times10^4\text{N}$
Here, -ve sign shows that force will be opposite to the direction of movement of the ball before hitting. So the force that the batsman had to apply to hold the bat firmly at its place would be F = 1.05 × 104N K