#### A 60 Kg woman consumes a 570 kcal chocolate cake for breakfast. If the human body is only 25% efficient in converting the chemical energy (food energy) into mechanical energy, how many steps will she have to climb to work off her breakfast ____. (Each step is 10 cm high and takes g= 9.8 m/ s^{2})

First of all, let’s represent chemical energy of the food consumed in terms of joule.

1 kcal = 4200 Joules

570 kcal = 2.394× 10^{6 }Joules

Now only 25% of this chemical energy is converted into mechanical energy therefore,

E_{mech } = 0.25×2.394× 10^{6} Joules
= 5.985 × 10^{ 5}joules

But E_{mech} = mgh, where m= mass of the woman=60Kg
h= total height of all steps

h = E_{mech}/mg = 5.985 × 10^{5}/ (60×9.8) = 1017.85 m

Height of each step is 0.1 m. Therefore, total no of steps that the woman has to climb = 1017.85/ 0.1≈ 10180 steps.

Note : The remaining chemical energy is converted into heat energy which the woman sweats out while climbing.

#### Which of the following is a non-conservative force?

Frictional force is not a conservative force as it depends on the path followed by the object and not independent of the path. The gravitational force, the electrostatic force and the magnetic force are path independent and depend only on the initial and final positions. Hence, these three forces are conservative forces.

#### A block of mass 0.15 kg is placed on a vertical spring (attached to a fixed end at the bottom) which is compressed to a certain length and then released so that the block rises to a height of 0.3m. The distance through which the spring is compressed is ____ (spring constant of spring is 100 N/m; g = 9.8 m/ s^{2}).

When the spring is compressed with the block on the top, work done by the spring kx^{2}/2 is stored as potential energy of the block mgh which then rises to a height h with that energy.

Therefore applying the law of conservation of energy,
kx^{2}/2= mgh or x^{2}=2mgh/k or x^{2}=(2×0.15×9.8×0.3)/100.

Hence x= 0.093 m ≈ 0.1 m.

Distance to which spring is compressed is 0.1 m

#### A man throws stones aiming at a building of height 14 m so that they just reach the top of the building. With what speed must he thrown the stones so that 40 % of the total energy is saved ____. (Acceleration due to gravity g= 9.8 m/ s^{2 })

The total energy associated with the stones = ½ mv^{2 } + mgh

Where, m - mass of the stones

v - velocity with stones are thrown

h - height of the building

When the stones just reach the top of the building, the man saves the kinetic energy. Therefore the percentage of energy saved is = [Kinetic energy/ total energy] × 100

Or (½ mv^{2} / [½ mv^{2} +mgh]) × 100 = 40
(v^{2}/ [v^{2} +2gh]) × 100 = 40
(v^{2}/[ v^{2} + 2 × 9.8× 14 ] ) = 0.4
v^{2} = 0.4 v^{2} +109.8
v ^{2}= 109.8 / 0.6 = 183

v = 13.52 m/s ≈ 14 m/s

#### A body of mass 5 kg is at rest. It is row subjected to a force of 18 N. The kinetic energy acquired by the body at the end of 25 s is ____.

a = F/m = 18/5 = 3.6 ms^{-1}

Therefore V = u + at = 0 + 3.6 × 25 = 90 ms^{-1}

Therefore, The kinetic energy acquired by the body.

K = 1/2 m V^{2}= 1/2×5× (90)^{2 }= 20250 J

#### The average horsepower (hp) required to raise a 100 Kg object through a height of 20 meters in 1 minute is ____.,

Average power in watts required for the purpose = mgh / time

P = (100 × 9.8 × 20) /60 Watts

P= 19600/60 = 326.6 Watts

1 Horse power = 746 Watts

Hence the required horsepower here is 0.43 hp

#### A block of mass m and a pan of equal mass are connected by a string going over a smooth pulley. Assume that the pulley is light. Initially, the system is at rest. A particle of mass m falls on the pan and sticks to it. If the particle strikes the pan with a speed v, the speed with which the system moves just after collision will be ____.

Let the required speed by V.

Let the impulse between the particle and the pan be J_{1}.

Let the impulse imparted to the block and the pan by the string be J_{2}.

Impulse = Change in momentum

For the particle, J_{1} = mv - mV ------- (1)

For the pan, J_{1} - J_{2} = mV ---------(2)

For the block, J_{2} = mV --------- (3)

Substituting (1) and (3) in (2)

(mv - mV) - mV = mV

mv = 3mV

V = v/3

#### Water falling down from a fall of height h is used for generating electrical energy. If 1.2 × 10^{4}kg of water falls per hour and if two thirds of the gravitational potential energy can be converted into electrical energy that can light fifty 100 W bulbs, the height h of the water fall is ____.

Mass of the falling water = m = 1.2×10^{4}kg/hr

Potential energy of falling water = mgh =1.2×10^{4}×9.8×h Joule/hr

Two thirds of the potential energy is converted into electrical energy therefore,

Electrical energy = 2/3 ×1.2×10^{4}×9.8×h Joule/hr

Power is defined as rate, at which energy is transferred therefore,

Power generated in one second = Electrical energy/ time = (2/3 ×1.2×10^{4}×9.8×h ) ÷ (3600) Watts or Joule/sec

This total power is used to light fifty 100 Watt bulbs.

1 bulb → 100 Watts

50 bulbs → (2/3 ×1.2×10^{4}×9.8×h) ÷ (3600) Watts

Therefore (2/3 ×1.2×10^{4}×9.8×h) ÷ (3600) = 50 ×100
Or h = ( 50 ×100×3×3600) ÷ (2×1.2×10^{4}×9.8) = 229.5 metres ≈230 metres.

Height of waterfall h = 230 metres.

#### A motorcyclist comes to a skidding halt in 5 m. In this process, the road exerts a force of 500 N. This force is opposed to the motion of the motorcycle. The work done by the road on the motorcycle is ____.

The work done by the frictional force that works towards bringing the motorcycle to a halt is the work done on the motorcycle by the road. The stopping force and the displacement of the motor cycle are in diagrammatically opposite directions. Thus, the angle between them is 180°.

Work done by the stopping force = Work done by the road (W_{r})

W_{r} = F s cosθ = 500 x 5 x cos 180° = 2500 J

#### The force constant of a weightless spring is 16 N/m. A body of mass 1 kg suspended from it is pulled down through 5 cm and then released. The maximum kinetic energy of the system (spring + body) will be ___ J.

According to the law of conservation of energy, the maximum kinetic energy of the system is equal to the maximum potential energy of the system.

Given, the force constant of the spring, k = 16 N/m

Mass of the body, m = 1 kg

Extension of the spring, x = 5 cm = 5×10^{–2}m

Energy of the compressed spring is given by

U=1/2kx^{2}=1/2×16×(5×10^{−2})^{2}= 0.02J = 2 × 10^{–2}J

#### A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0 m. How much energy is dissipated by air drag acting on the ball during the ascent?

When the ball is thrown up, the kinetic energy of the ball is converted to its potential energy.

Kinetic energy of the ball when
projected=1/2mv^{2}=1/2×0.5×14^{2}=49 J

Given the height gained by the ball, h = 8.0 m

When the ball reaches its maximum height, the total energy of the ball is its potential energy = mgh = 0.5 × 9.8 × 8 = 39.2 J

As observed the potential energy gained by the ball at its position of maximum height is less than its kinetic energy at the point of projection.

According to the law of conservation of energy, the total energy is constant, which implies that the difference of the kinetic and potential energies of the ball is equal to the energy dissipated by the air drag on the ball during its ascent.

Hence, the energy dissipated by air drag on the ball during its ascent = 49− 39.2 = 9.8 J

#### Which of the following statements is incorrect? In an inelastic collision, ___.

In an inelastic collision, linear momentum is conserved. However, in an elastic collision, apart from linear momentum, kinetic energy is also conserved.

#### A spring gun of spring constant 90 N/cm is compressed 12 cm by a ball of mass 16 g. If the trigger is pulled, the velocity of the ball is ____ m/s.

Given, spring constant, k = 90 N/cm =90N/1cm=90N/0.01m = 9000 N/m

Compression of the spring, x = 12 cm = 0.12 m

Mass of the ball, m = 16 g = 0.016 kg

The mechanical energy of the system is conserved. Here, the loss in potential energy of the spring is equal to the gain in kinetic energy of the ball.

1/2kx^{2}=1/2mv^{2}, where ‘v’ is the velocity of the ball.

⇒ mv^{2}=kx^{2}

⇒ v2=kx^{2}/m

⇒v=kx^{2}/m

=√ [(9000)×( 0.12)^{2}]/0.016

= 90 m/s

#### A spring of force constant 500 N/m has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is ___ J.

Given the force constant of the spring, k = 500 N/m

Initial extension of the spring, x_{1} = 5 cm = 0.05 m

Final extension of the spring, x_{2} = 15 cm = 0.15 m

Work done in extending a spring from “x_{1}” to “x_{2}”is given by:

W=1/2k(X2/2−X2/1)=1/2×500×(0.15^{2}−0.05^{2})=5 J

#### A mason carries a cement bag. The force applied by him to carry this is 5N. He carries this bag over a horizontal distance of 10 m and a vertical height of 15 m. The total work done by the mason is ____.

When the mason travels the horizontal distance, the angle between force and displacement is 90º.

In this case, F = 5 N, s = 10 m and θ = 90º.

Work done while moving in the horizontal direction (W1) = Fs × cosθ = 5 × 10 × cos 90º = 0 (cos 90 = 0)

When the mason travels the vertical distance, the angle between force and displacement is 0º.

In this case, F = 5 N, s = 15 m and θ = 0º.

Work done while moving in the vertical direction (W2) = Fs × cosθ = 5 × 15 × cos 0º = 75 J (cos 0 = 1)

Total work done = W_{1 }+ W_{2} = 0 + 75 J

#### Two blocks of masses m1 = 3 kg and m_{2} = 4 kg are connected by a light horizontal spring (shown in the image). They lie on a rough horizontal surface. The coefficient of friction between the blocks and the surface is µ= 0.5. What minimum constant force F has to be applied in the horizontal direction to the block of mass m1 in order to shift the other blocks ____. (Use g = 9.8 m/ s^{2}).

From the work-energy theorem, we know that net work done = change in kinetic energy

Net work done = Fx− kx^{2}/2− μm_{1}gx where Fx is the work done by the external force to move block m_{1} by a distance x m

kx^{2}/2 is the work done by the spring to compress the spring by x m μm1gx is the work done against friction

The system starts from rest and finally comes to rest; therefore change in kinetic energy = 0

Fx− kx^{2}/2− μm_{1}gx = 0 -------- (1)

But in order to move m_{2} the spring force kx must be exactly equal to the force of friction that acts against the motion of m_{2 } = µm_{2}g

Therefore, kx= μm_{2}g --------- (2)

Using (2) in (1), Fx −μm_{2}g x/2 -μ m_{1}gx = 0 or F =μm_{1}+m_{2}/2g

Here, m_{1} = 3 kg, m_{2} = 4 kg , μ =0.5 , g = 9.8 m/s^{2 }

Thus we obtain F = 24.5 N.

#### A rubber ball is released from certain height. It loses 50% of its kinetic energy on striking the ground and bounces. On bouncing, it attains a height equal to

When the rubber ball is just released, its energy is completely potential and is given by “mgh” where ‘h’ is the height from which the ball is released. On its transit to the ground, its potential energy is gradually converted to its kinetic energy. On striking the ground, 50% of its energy is lost. Hence, the remaining energy is 50% and when bounced this 50% of the energy is converted to potential energy. Therefore, the height to which the rubber ball rises is only half of the initial height.

#### Which of the following is not an example of work done by a variable force?

Work done in lifting a load by a crane is not an example of work done by a variable force. This is because; the force in lifting a load by a crane is not variable. It is constant and is equal to the weight of the load. In the other three cases, the force is variable.

#### The disintegration energy Q for the fission reaction of ^{98} Mo ^{98}Mo→ ^{49}Sc + ^{49}Sc is ____, (Given mass of ^{98}Mo → m Mo = 97.905 atomic mass units (amu), mass of ^{49} Sc → m_{Sc} = 48.9502 amu)

Disintegration energy Q = (m _{Mo} –2 m _{Sc}) c^{2}

But c^{2} = 931.5 MeV/amu

Q = (97.905 amu – 2× 48.9502 amu ) × 931.5 MeV/amu

Q = 4.28 MeV

#### A particle is moving under the influence of a force given by F = kx where “k” is a constant and “x” is the displacement. The energy in joule, gained by the particle in moving from x = 0 to x = 4 is:

Energy in moving the particle from “x_{1}” to “x_{2}” is given
by U=1/2k(x^{2}_{2}−X^{2}_{1}), where “k” is the constant

Hence, U=1/2k(4^{2}−0^{2})=8k

#### The decrease in potential energy of a body of mass 30 kg which falls from a height of 50 cm is ___ J.

Decrease in potential energy of a body when it falls through a height Δh is given by "mgΔh", where ‘m’ is the mass of the body and ‘g’ is acceleration due to gravity.

Given, mass of the body, m = 30 kg

Acceleration due to gravity g = 9.8 m/s^{2}

Decrease in the height of the body, Δh = 50 cm = 0.5 m

Hence, decrease in potential energy of the body, ΔU = mgΔh =30×9.8× 0.5 = 147 J

#### When two objects are in direct head-on impact, the speed with which they separate after impact is ____.

It has been observed experimentally that the ratio of the separation speed and approach speed is constant for two given set of bodies. This property formulated by Newton is known as the law of restitution. This ratio e is called the coefficient of restitution.

#### In which case does the potential energy decrease?

In compressing a spring, work is done on the body, which is the spring. Work is done against the restoring force and hence the potential energy of the spring increases. Same is the case when a spring is stretched.

On moving a body against the gravitational force, the work done is stored in it in the form of potential energy. Thus, in the case of a body moving against gravitational force the potential energy of the body increases.

In the case of rising of an air bubble in water, the work is done in the direction of the up thrust. Hence, the potential energy of the system decreases.

#### An overhead tank at a height of 500 m above the ground level burst and the water fell over an area of 50 m^{2}. Water settled in a puddle of uniform depth of 5 cm. The work that would have gone in raising water to the overhead tank is ____.

Volume of water fallen on the ground = A × Depth

Area of water A = 50 m^{2}; Depth of water = 5 cm

Therefore, volume of water = 50 × 5 × 10^{-2}= 2.5 m^{3}

Mass of water m = Volume × Density = 2.5 × 10^{3} kg

Force used in raising water to the overhead tank = mg = 2.5 × 10^{3} × 9.8 N

Work done = Fs = 2.5 × 10^{3} × 9.8 × 0.5 = 12250 J

#### If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is ____ m/s(Take g = 9.8m/s^{2})

The potential energy of the water at a higher level is converted to its kinetic energy on falling on to the turbine. Potential energy = mgh

Kinetic energy =1/2m v^{2}

mgh=1/2m v^{2}

⇒ v^{2} = 2gh

v=√(2gh)

=√(2×9.8×19.6)

=19.6 m/s

#### A block of mass 0.35 kg moves at a speed of 5 m/s and compresses a spring through a distance of 0.08 m, after which its speed is reduced to half the initial value. The spring constant of the spring is ___.

Initially the block will possess only kinetic energy. When it comes into contact with the spring and compresses it, a portion of the kinetic energy is converted to the potential energy of the spring and the rest is the kinetic energy of the block that bounces back.

Hence, initial energy = m v_{i}^{2} / 2, where vi is the initial speed of the block.

Final energy=
mv_{f}^{2}/2+ kx^{2}/2... , where vf is the final speed and k the spring constant. x is the distance through which the spring is compressed.

From the principle of conservation of energy, m v_{i}^{2}/2 = mv_{f}^{2}/2+kx^{2}/2... Or kx^{2} = m (v_{i}^{2}-v_{f}^{2})

k =0.35×5^{2}/0.08^{2}

− 0.35×2.5^{2/}0.08^{2} ≈ 1025 N/m

#### A small ball of mass m is placed on top of a 'super ball' of mass M. Both balls are dropped to the floor from a height h. How high does the small ball rise after collision? (You can assume that the collision between the balls is elastic and that the mass of the small ball is far lesser than the mass of the big ball.)

When the super ball hits the floor, the balls are slightly separated.

Let v1 be the velocity of the small ball, just after collision..

We know that v_{1}=[( m_{1}− m_{2})]/ m_{1}+m_{2})× v_{1}+(2m_{2}/m_{1}+m_{2})×v_{2}
v_{1} = v_{0 }and v2 = –v0

m_{1}/m_{2} =m/M=0

Therefore, v_{1}' = -3 v_{0 } = -3√2gh

Therefore, v_{1}=−3v_{0}=−3√2gh

Therefore, height gained by the small ball after collision =v1=9(2gh)/2g= 9h

#### The displacement of a body of mass 3 Kg varies with time as S = t^{2}+ 2, where S is in meters and t is in seconds. The Work done by all the forces acting on the body during the time interval t = 3 s to t = 6 s is ____.

Velocity is defined as rate of change of displacement of the particle. Therefore, v = dS/dt = d/dt(t^{2}+2) = 2t m/s
From Work- Energy theorem;
The total Work done on a particle equals the change in its Kinetic energy (KE);

W_{net} = Change in KE = Final KE− Initial KE = K_{f }− K_{i } K_{f} = 1/2×m×v_{f}^{2}, where m is the mass of the body and vf is the final velocity i.e., velocity at t = 6 s.
v_{f } = 2t at t = 6 s→ v_{f } = 12 m/s

Therefore Kf = 1/2×3×12^{2} = 216 Joules

Similarly K_{i} = 1/2 ×m×v_{i}^{2}, where v_{i} is the initial velocity i.e., velocity at t = 3s
v_{i } = 2t at t = 3 s→ v_{i } = 6 m/s

Therefore K_{i} = 1/2×3×6^{2}= 54 Joules

Hence Total Work done by all forces on the body during time interval t = 3 s to t = 6 s is W_{net} = 216−54 = 162 Joules.

#### The potential energy of a spring when stretched through length “x” is 20 J. The amount of work that must be done in order to stretch it through an additional length of “x” will be ___ J

Potential energy of the spring is given by
U=1/2 kx^{2}, where “k” is the spring constant and “x” is the extension.

Given 20=1/2kx^{2} ---- (1)

Let U' be the work done in stretching the spring through an additional length “x”.

Then the total extension is “2x”

U'=1/2k(2x)^{2}=1/2k(4x^{2})

=4(1/2kx^{2})

= 4 × 20 = 80 J

#### A cart man pushes a cart through a distance of 50 m. If he applies a force of 15 kg wt in a direction inclined at 60°to the ground, the work done by him is ____ (Assume g = 9.8 ms^{-2})

F = 15 kg wt = 15 × 9.8 N

s = 50 m, θ = 60º

We know that W = F s cosθ

W = 15 × 9.8 x 50 × cos 60°

= 15 × 9.8 × 50 x ½

= 3675 J

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