Work Power & Energy — Page 5 | NEET Physics

Q41

An engine pumps water continuously through a hose. Water leaves the hose with a velocity v and m is the mass per unit length of the water jet. What is the rate at which kinetic energy is imparted to water?

A mv 3

B

{1 \over 2}m{v^2}

C

{1 \over 2}{m^2}{v^2}

D

{1 \over 2}m{v^3}

Correct Answer

Option D

Solution

Let m be the mass per unit length, so the rate of mass leaving the hose per sec = mv Rate of kinetic energy K.E =

{1 \over 2}\left( {mv} \right){v^2} = {1 \over 2}m{v^3}

Q42

Water falls from a height of 60 m at the rate of 15 kg/s to operate a turbine. The losses due to frictional forces are 10% of energy. How much power is generated by the turbine ? (g = 10 m/s 2 )

A 12.3 kW

B 7.0 kW

C 8.1 kW

D 10.2 kW

Correct Answer

Option C

Solution

It is seen that input power to turbine will be the potential energy due to water falling in a second which is equal to mgh = 15×10×60 = 9000 watt Now power generated by turbine is 10% less than above value equal to (9000 – 900) watt = 8100 watt = 8.1 kW

Q43

A vertical spring with force constant k is fixed on a table. A ball of mass m at a height h above the free upper end of the spring falls vertically on the spring so that the spring is compressed by a distance d. The net work done in the process is

A mg(h + d)

-

{1 \over 2}

kd 2

B mg(h

-

d)

-

{1 \over 2}

kd 2

C mg(h

-

d) +

{1 \over 2}

kd 2

D mg(h + d) +

{1 \over 2}

kd 2

Correct Answer

Option A

Solution

As seen, gravitational potential energy of the ball gets converted into elastic potential energy of the spring.

mg\left( {h + d} \right) = {1 \over 2}k{d^2}

W net = work done by gravity + work done by spring $\Rightarrow$ W net = mg(h + d) -

{1 \over 2}k{d^2}

Q44

A body of mass 3 kg is under a constant force which causes a displacement s in metres in it, given by the relation s =

{1 \over 3}

t 2 , where t is in seconds. Work done by the force in 2 seconds is

A

{{19} \over 5}J

B

{5 \over {19}}J

C

{3 \over {8}}J

D

{8 \over {3}}J

Correct Answer

Option D

Solution

Acceleration = d 2 s/dt 2 = 2/3 m/s 2 Now force acting on body = mass × acceleration F = 3 × (2/3) = 2 Newton Hence displacement in 2 secs = (1/3) × 2 × 2 = 4/3 m Finally, work done = 2 × 4/3 = 8/3 J

Q45

The potential energy of a long spring when stretched by 2 cm is U. If the spring is stretched by 8 cm the potential energy stored in it is

A U/4

B 4U

C 8U

D 16U

Correct Answer

Option D

Solution

Potential energy of a spring =

{1 \over 2}

$\times$ force constant $\times$ (extension) 2 $\therefore$ Potential energy $\propto$ (extension) 2 or,

{{{U_1}} \over {{U_2}}} = {\left( {{{{x_1}} \over {{x_2}}}} \right)^2}

or,

{{{U_1}} \over {{U_2}}} = {\left( {{2 \over 8}} \right)^2}

\Rightarrow {{{U_1}} \over {{U_2}}} = {1 \over {16}}

\Rightarrow {U_2} = 16{U_1} = 16U\,\,\left( {\because{U_1} = U} \right)

Q46

300 J of work is done in sliding a 2 kg block up an inclined plane of height 10 m. Work done against friction is (Take g = 10 m/s 2 )

A 1000 J

B 200 J

C 100 J

D zero

Correct Answer

Option C

Solution

Loss in potential energy = mgh = 2 × 10 × 10 = 200 J.

Gain in kinetic energy = work done = 300 J $\therefore$ Work done against friction = 300 – 200 = 100 J

Q47

A force F acting on an object varies with distance x as shown here. The force is in N and x in m. The work done by the force in moving the object from x = 0 to x = 6 m is

A 18.0 J

B 13.5 J

C 9.0 J

D 4.5 J

Correct Answer

Option B

Solution

Work done = area under F-x graph = area of trapezium OABC =

{1 \over 2}\left( {3 + 6} \right)\left( 3 \right) = 13.5\,J

Q48

A mass of 0.5 kg moving with a speed of 1.5 m/s on horizontal smooth surface, collides with a nearly weightless spring of force constant k = 50 N/m. The maximum compression of the spring would be

A 0.15 m

B 0.12 m

C 1.5 m

D 0.5 m

Correct Answer

Option A

Solution

It is observed that if mass colloids with spring in such case, spring will be compressed, so kinetic energy of mass will be equal to that of springs potential energy.

Further, if y is the compression, then

{1 \over 2}m{v^2} = {1 \over 2}k{y^2}

\Rightarrow {1 \over 2}\left( {0.5} \right)\left( {1.5} \right) = {1 \over 2}\left( {50} \right){y^2}

$\therefore$ y = 0.15 m

Q49

A particle of mass m 1 is moving with a velocity v 1 and another particle of mass m 2 is moving with a velocity v 2 . Both of them have the same momentum but their different kinetic energies are E 1 and E 2 respectively. If m 1 > m 2 then :

A E 1 < E 2

B

{{{E_1}} \over {{E_2}}} = {{{m_1}} \over {{m_2}}}

C E 1 > E 2

D E 1 = E 2

Correct Answer

Option A

Solution

Kinetic energy =

{{{p^2}} \over {2m}}

{{{E_1}} \over {{E_2}}} = {{p_1^2/2{m_1}} \over {p_2^2/2{m_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}

as m 1 > m 2 $\therefore$ E 1 < E 2

Q50

A ball of mass 2 kg and another of mass 4 kg are dropped together from a 60 feet ball building. After a fall of 30 feet each towards earth, their respective kinetic energies will be in the ratio of

A

\sqrt 2

: 1

B 1 : 4

C 1 : 2

D 1 :

\sqrt 2

Correct Answer

Option C

Solution

From the question, as height is same for both balls, so their velocities on reaching the ground will be same.

Hence in such case kinetic energy ∝ mass So,

{{K.{E_1}} \over {K.{E_2}}} = {{{m_1}} \over {{m_2}}} = {1 \over 2}