Work Power And Energy(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

WORK POWER AND ENERGY MCQs

Total Questions : 30 | Page 1 of 3 pages

A moving railway compartment has a spring of constant k fixed to its front wall. A boy stretches this spring by distance x and in the mean time the compartment moves by a distance s. The work done by boy w.r.t. earth is

$\frac{1}{2} k x^{2}$
$\frac{1}{2} (k x) (s + x)$
$\frac{1}{2} k x s$
$\frac{1}{2} k x (s + x + s)$

Discuss Question

Answer: Option A. ->

\frac{1}{2} k x^{2}

:
A

Total work done is equal to the energy stored in the spring.

Question 2.

A block of mass 5.0 kg slides down from the top of an inclined plane of length 3 m. The first 1 m of the plane is smooth and the next 2 m is rough. The block is released from rest and again comes to rest at the bottom of the plane. If the plane is inclined at $30^{\circ}$ with the horizontal, find the coefficient of friction on the rough portion.

$\frac{2}{\sqrt{3}}$
$\frac{\sqrt{3}}{2}$
$\frac{\sqrt{3}}{4}$
$\frac{\sqrt{3}}{5}$

Discuss Question

Answer: Option B. ->

\frac{\sqrt{3}}{2}

:
B

Force Method:

Form P to Q: $v^{2} = O^{2} + 2 a_{1} S_{1}$

Where $a_{1} = g s i n 30^{\circ}, s_{1} = 1 m$

And form Q to R: $O^{2} = v^{2} + 2 a_{2} s_{2}$

Where $a_{2} = g s i n 30^{\circ} - μ g c o s 30^{\circ}, s_{2} = 2 m$

Solve to get $μ = \frac{\sqrt{3}}{2}$

Question 3.

Given $⃗ F = (x y^{2})^i + (x^{2} y)^j N$ . The work done by $⃗ F$ when a particle is taken along the semicircular path OAB where the coordinates of B are (4,0) is

$\frac{65}{3} J$
$\frac{75}{2} J$
$\frac{73}{4} J$
Zero

Discuss Question

Answer: Option D. -> Zero
:
D

Given $⃗ F = (x y^{2})^i + (x^{2} y)^j$

W = $\int F_{x} d x + \int F_{y} d y$

= $\int x y^{2} d x + \int x^{2} y d y$

= $\frac{1}{2} \int d (x^{2} y^{2}) = {[\frac{x^{2} y^{2}}{2}]}_{(0, 0)}^{(4, 0)} = 0$

Question 4.

A small block of mass 2 kg is kept on a rough inclined surface of inclination $θ = 30^{\circ}$ fixed in a lift. The lift goes up with a uniform speed of 1 m $s^{- 1}$ and the block does not slide relative to the inclined surface. The work done by the force of friction on the block in a time interval of 2 s is.

Zero
9.8 J
29.4 J
16.9 J

Discuss Question

Answer: Option B. -> 9.8 J
:
B

Since the lift's speed is constant, in the absence of acceleration, there will be no pseudo-force (referring to the lift as the frame of reference) or additional reaction/thrust due to inclined plane (referring to ground as the reference frame) on the particle.
Therefore, friction =mg sin $θ$ acting along the plane.
Distance moved by the particle (or lift) in time t=vt

Work done in time t=(mg sin $θ$ )vt(cos $(90^{\circ}$ - $θ$ ))=mg $s i n^{2} θ$ vt

Substituting the Values we get Work = 9.8 J

Question 5.

A particle of mass m moves with a variable velocity v, which changes with distance covered x along a straight line as $v = k \sqrt{x}$ ,where k is a positive constant . The work done by all the forces acting on the particle, during the first t second is

$\frac{m k^{4}}{t^{2}}$
$\frac{m k^{4} t^{2}}{4}$
$\frac{m k^{4} t^{2}}{8}$
$\frac{m k^{4} t^{2}}{16}$

Discuss Question

Answer: Option C. ->

\frac{m k^{4} t^{2}}{8}

:
C

GIven $v = k \sqrt{x}$ or $\frac{d x}{d t} = k \sqrt{x}$ or $x^{- \frac{1}{2}} d x = k d t$

Integrating both sides, we get

$\frac{x^{\frac{1}{2}}}{\frac{1}{2}} = k t + C$ ; Assuming x(0) = 0

Therefore, C = 0

$2 \sqrt{x} = k t \Rightarrow x = \frac{k^{2} t^{2}}{4}$ or $v = \frac{k^{2} t}{2}$

Therefore, work done,

$△ W$ = Increase in KE

= $\frac{1}{2} m v^{2} - \frac{1}{2} m (0)^{2} = \frac{1}{2} m {[\frac{k^{2} t}{2}]}^{2} = \frac{m k^{4} t^{2}}{8}$

Question 6.

What is the angle between the following vectors?

$⃗ A =^i +^j +^k$ and $⃗ B = - 2^i - 2^j - 2^k$ .

$45^{\circ}$
$90^{\circ}$
$0^{\circ}$
$180^{\circ}$

Discuss Question

Answer: Option D. ->

180^{\circ}

:
D

$⃗ A . ⃗ B = | A | | B | c o s θ o r, c o s θ = \frac{⃗ A . ⃗ B}{| A | | B |}$ ----------------(i)

But, $⃗ A . ⃗ B = (^i +^j +^k) . (- 2^i - 2^j - 2^k)$

$⃗ A . ⃗ B = - 2 - 2 - 2$

$⃗ A . ⃗ B = - 2 - 2 - 2 = - 6$

Again A = $| ⃗ A | = \sqrt{(1)^{2} + (1)^{2} + (1)^{2}} = \sqrt{3}$ ;

B = $| ⃗ B | = \sqrt{(- 2)^{2} + (- 2)^{2} + (- 2)^{2}} = \sqrt{12} = 2 \sqrt{3}$

Now, $c o s θ = \frac{- 6}{\sqrt{3} \times 2 \sqrt{3}} = - 1 \Rightarrow θ = 180^{\circ}$

Question 7.

What is the angle between the following vectors?

$\to A = 3^i - 2^j +^k$

$\to B = 2^i + 6^j - 6^k$

$c o s^{- 1} (\frac{12}{\sqrt{266}})$
$c o s^{- 1} (\frac{- 6}{\sqrt{266}})$
$c o s^{- 1} (\frac{- 6}{\sqrt{1064}})$
None of these

Discuss Question

Answer: Option B. ->

c o s^{- 1} (\frac{- 6}{\sqrt{266}})

:
B

$\to A . \to B = (3 \times 2) + (- 2 \times 6) + (1 \times - 6)$

= 6 - 12 - 6

= -12

$| \to A | = \sqrt{3^{2} + (- 2)^{2} + 1^{2}} = \sqrt{9 + 4 + 1} = \sqrt{14}$

$| \to B | = \sqrt{2^{2} + 6^{2} + (- 6)^{2}} = \sqrt{4 + 36 + 36} = \sqrt{76}$

$\frac{\to A . \to B}{| \to A | | \to B |} = \frac{- 12}{\sqrt{14} \sqrt{76}} = \frac{- 6}{\sqrt{266}}$

$∴ c o s θ = \frac{- 6}{\sqrt{266}}$

$\Rightarrow θ = c o s^{- 1} (\frac{- 6}{\sqrt{266}})$

Question 8.

$⃗ F = 3^i + 6^j - 8^k$

$⃗ s = - 2^i + 5^j + 2^k$

If W = $⃗ F . ⃗ s$ find W.

8 N. m
20 N. m
5 N. m
None of these

Discuss Question

Answer: Option A. -> 8 N. m
:
A

$⃗ F = 3^i + 6^j - 8^k$

$⃗ s = - 2^i + 5^j + 2^k$

W = $⃗ F . ⃗ s$

= $(3^i + 6^j - 8^k) . (- 2^i + 5^j + 2^k)$

= (-6) + (30) + (-16)

$∵ (^i .^i = 1 &^i .^j = 0)$

W = 8 N.m

Question 9.

$| \to A | = 4 u n i t s$

$| \to B | = 10 u n i t s$ If angle between $\to A$ & $\to B$ is $60^{\circ}$ then find $\to A$ . $\to B$ .

$20$
$40$
$0$
$20 \sqrt{3}$

Discuss Question

Answer: Option A. ->

20

:
A

$\to A$ . $\to B$ = $| A | | B | c o s 60^{\circ}$ .

$| ¯ ¯¯ ¯ A | = 4$ ; $| ¯ ¯¯ ¯ B | = 10$

= $4 \times 10 \times \frac{1}{2}$

Scalar product of $\to A$ & $\to B$ = 20

Question 10.

A force $\to F = 3^i + 2^j + c^k$ acts on a particle and causes a displacement $\to r = c^i + 4^j + c^k$ meter.

The work done is 36 joule.Find the value of 'c' ?

7
-7
4
0

Discuss Question

Answer: Option B. -> -7
:
B and C

$\to F = 3^i + 2^j + c^k$

$\to r = c^i + 4^j + c^k$

Work = 36j

W = $\to F . \to r$

36 = $(3^i + 2^j + c^k) . \to F = (c^i + 4^j + c^k)$

36 = 3c + 8 + $c^{2}$

$\Rightarrow c^{2} + 3 c - 28 = 0$

(c+7) (c - 4)

c = -7 or 4

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