If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of ...

[Laws of Motion]

If two forces of 5 N each are acting along X and Y axes, then the magnitude and direction of resultant is

1) 5√2,π/3 2) 5√2,π/4 3) −5√2, π/3 4) −5√2,π/4

Correct Answer :2

SOLUTION:

And tanā”θ=5/5 = 1, => θ =π /4

A body of mass 2kg is projected vertically upwards with speed of 3m/s. The maximum gravitational ...

[Work Power Energy]

A body of mass 2kg is projected vertically upwards with speed of 3m/s. The maximum gravitational potential energy of the body is

1) 18J 2) 4.5J 3) 9 J 4) 2.25J

Correct Answer :3

**SOLUTION**

At maximum height velocity is zero thus Kinetic energy=0 and Potential energy max

ΔK.E = -ΔP.E.

Initial KE = ½ m v^{2}

Initial KE = ½ 2 (3)^{2}

Initial kinetic energy = 9 J

The heart of a man pumps 4000 cc of blood through the arteries per minute at a pressure of 130 ...

[Work Power Energy]

The heart of a man pumps 4000 cc of blood through the arteries per minute at a pressure of 130 Hg. If the density of mercury is 13.6 ×10^{3} kg/m^{3}, what is the power of heart ?

1) 1.15 watt 2) 2.30 watt 3) 1.15 hp 4) 0.115 hp

Correct Answer :1

SOLUTION:−

**SOLUTION**

We know that power is equal to rate of doing work. In the given configuration, we can consider that the given volume of blood has been raised up by a certain height, whose equivalent height in terms of mercury is 130 mm. Thus power can be calculated by equating it to rate of increase of potential energy of the equivalent mercury column (instead of blood column). Thus calculation can be done by using the formula, P=mgh / t

Now, here mass of 4000 c.c. of mercury should be calculated, as in equivalent terms, the work done in pushing 4000 c.c.of blood to a certain height which is not given in the problem is same as the work done in pushing the same volume of mercury to a height of 130 mm.

An open knife edge of mass M is dropped from a height h on a wooden floor. If the blade ...

[Work Power Energy]

An open knife edge of mass M is dropped from a height h on a wooden floor. If the blade penetrates S into the wood, the average resistance offered by the wood to the blade is

1) Mg 2) Mg(1+h / s) 3) Mg(1+hS) 4)

Correct Answer :2

SOLUTION:−

Let the average resistive force offered be F then,

work done by F will be F ∗ S

work done by gravity =m ∗ g ∗ (h + S)

Now , when the knife was released it's velocity was zero and finally also it came to rest .

Therefore the net work done on the knife should be zero

F∗S=m∗g∗(h + S)

F=m∗g∗(1+h / S)

A spring with spring constant K when stretched through 2cm the potential energyis U. If it is ...

[Work Power Energy]

A spring with spring constant K when stretched through 2cm the potential energyis U. If it is stretched by 6cm. The potential energy will be

1) 6U 2) 3U 3) 9U 4) 18U

Correct Answer :3

SOLUTION:−

Potential energy of spring U ∝ X^{2}

Thus U'/U=(6/2)^{2}

U'/U=3^{2}

U'=9U

A spring of force constant 10 N per meter has an initial stretch 0.20 m. In changing the stretch ...

[Work Power Energy]

A spring of force constant 10 N per meter has an initial stretch 0.20 m. In changing the stretch to 0.25m, the increase in potential energy is about

1) 0.1 joule 2) 0.2 joule 3) 0.3 joule 4) 0.5 joule

Correct Answer :1

SOLUTION:−

Elastic potential energy of spring = ½ kx^{2}

Change in energy = ½10 ( 0.25^{2} - 0.20^{2})

Change in energy = 0.1125 J

A mass M is attached to a massless spring of spring constant K. The mass is moved through a ...

[Work Power Energy]

A mass M is attached to a massless spring of spring constant K. The mass is moved through a distance x and released. The maximum velocity of the mass is:

1) 2πM / K 2) 3) √ K /M 4) √ K / Mx

Correct Answer :2

SOLUTION:−

Initial total energy of the system = KE + PE

initially KE is zero and PE is (k*x^{2})/2

as we know, maximum velocity will be attained when the block is at the equilibrium position because at equilibrium net force on the block will be zero.

now, when the block passes through the equilibrium position

we have,

PE is zero as there is neither expansion nor contraction in the spring

KE=(m*v^{2})/2

applying conservation of energy gives us

(m*v^{2})/2=(k*x^{2})/2

A chain is held on a friction-less table with one third of its length hanging over the edge. The ...

[Work Power Energy]

A chain is held on a friction-less table with one third of its length hanging over the edge. The total length of the chain is 1 and its mass is m. Find the work required to pull the hanging part back to the table.

1) 2mgl 2) mgl / 5 3) mgl / 9 4) mgl / 18

Correct Answer :4

SOLUTION:−

The mass of hanging chain is m / 3, length is l / 3 .

Now the the center of mass of hanging chain is at a distance of l / 6 from table top.

The mass hanging over the edge is mg / 6, so initial force required to pull the hanging mass on the table =mg / 3.

The final force required to pull the remaining mass = 0.

So, Average force =mg / 6

Displacement in the chain = l / 3.

Thus, work done by the variable force

= force x distance =

Where 1 / x is the fraction of the length of the chain which is initially hanging

Here, x = 3

So, work done =

A block of mass 10 kg, moving in x direction with a constant speed of 10ms^{-1}, is ...

[Work Power Energy]

A block of mass 10 kg, moving in x direction with a constant speed of 10ms^{-1}, is subjected to a retarding force F = 0.1 x J/m during its travel from x = 20 m to 30 m. Its final KE will be :

1) 250 J 2) 475 J 3) 450 J 4) 275 J

Correct Answer :2

**SOLUTION **

**Initial K.E. = ½ mv ^{2} = 500 J
Work done by retarding force, retarding force is position dependent
**

**Final K.E. = Initial K.E. – Work done by friction
Final K.E. = 500 – 25 = 475 J**