Work Energy and Power Online Objective Test | Work Energy and Power Online Objective Test

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q1.

The Sun converts an enormous amount of matter to energy. Each second, 4.19 $\times$ $1 0^{9}$ kg—approximately the capacity of 400 average-sized cargo ships—is changed to energy. What is the power output of the Sun?

\times

2.62

\times

1 0^{26}

B. 1.57

\times

1 0^{26}

0.72

\times

1 0^{26}

D. 3.77×1026W

Answer : Option D
Explaination / Solution:

Energy liberated per second

$E = m c^{2} = 4.19 \times 10^{9} \times 3 \times 10^{8} \times 3 \times 10^{8} = 37.71 \times 10^{25} J$

power output of sun is equal to energy output per second

$P = \frac{W}{t} = \frac{37.71 \times 10^{25}}{1} = 3.77 \times 10^{26} W$

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q2. A block having a mass of 0.80 kg is given an initial velocity 1.2 m/s to the right and collides with a spring of negligible mass and force constant k =50 N/m. Assuming the surface to be frictionless, calculate the maximum compression of the spring after the collision.

A. 0.25 m

B. 0.10 m

C. 0.15m

D. 0.20 m

Answer : Option C
Explaination / Solution:

for maximum compression

$\begin{aligned} \frac{1}{2} k x^{2} = \frac{1}{2} m v^{2} \\ x = \sqrt{\frac{m v^{2}}{k}} = \sqrt{\frac{0.8 \times 1.2 \times 1.2}{50}} = 0.15 m \end{aligned}$

the maximum compression of the spring after the collision = 0.15m

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q3. The launching mechanism of a toy gun consists of a spring of unknown spring constant. When the spring is compressed 0.120 m, the gun, when fired vertically, is able to launch a 35.0-g projectile to a maximum height of 20.0 m above the position of the projectile before firing. Neglecting all resistive forces, determine the spring constant.

A. 953 N/m

B. 903 N/m

C. 873 N/m

D. 993 N/m

Answer : Option A
Explaination / Solution:

Potential energy of spring converted in to potential energy

$\begin{aligned} \frac{1}{2} k x^{2} = m g h \\ k = \frac{2 m g h}{x^{2}} = \frac{2 \times 35 \times 10^{- 3} \times 9.8 \times 20}{0.12 \times 0.12} = 953 N / m \end{aligned}$

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q4. A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m/s. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact?

A. 8.42 J

B. 8.82 J

C. 8.11 J

D. 9.22 J

Answer : Option B
Explaination / Solution:

Whole of the potential energy of bolt converted in to heat energy heat produced by the impact = mgh =0.3×9.8×3=8.82J

$\begin{aligned} m g \sin 37^{\circ} - μ m g \cos 37^{\circ} = \frac{1}{2} k x^{2} \\ (1 \times 9.8 \times 0.6) - (μ \times 1 \times 9.8 \times 0.8) = \frac{1}{2} \times 100 \times 0.1 \\ 0.602 - 0.799 μ = 0.510 \\ μ = \frac{0.092}{0.799} = 0.115 \end{aligned}$

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q5.

A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m $s^{- 1}$ relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley?

A. 11.36 m/s

B. 8.13 m/s

C. 9.36 m/s

D. 10.36 m/s

Answer : Option D
Explaination / Solution:

Mass of troly M = 200Kg

mass of child m = 20Kg

speed of trolley v = 36Km/hr=36 x 5/18 = 10m/s

Let v' be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground = $v^{,} - 4$

from conservation of linear momentum

$\begin{aligned} p_{i} = p_{f} \\ (M + m) v = M v^{,} + m (v^{,} - 4) \\ (200 + 20) \times 10 = 200 v^{,} + 20 (v^{,} - 4) \\ 2200 = 220 v^{,} - 80 \\ v^{,} = \frac{2280}{220} = 10.36 m / s \end{aligned}$

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q6. A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m

s^{- 1}

relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. How much has the trolley moved from the time the child begins to run?

A. 25.9 m

B. 24.1 m

C. 27.8 m

D. 23.3 m

Answer : Option A
Explaination / Solution:

Mass of troly M = 200Kg

mass of child m = 20Kg

speed of trolley v = 36Km/hr=36 x 5/18 = 10m/s

Let v' be the final velocity of the trolley with respect to the ground.

Final velocity of the boy with respect to the ground = $v^{,} - 4$

from conservation of linear momentum

Time taken by the boy to run t = 10/4 = 2.5 sec

Distance moved by the trolley = $v^{,} t = 10.36 \times 2.5 = 25.9 m$

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q7. A 50.0-kg marathon runner runs up the stairs to the top of a 443-m-tall Tower. To lift herself to the top in 15.0 minutes, what must be her average power output?

A. 241 W

B. 221 W

C. 201 W

D. 261 W

Answer : Option A
Explaination / Solution:

P=WtW=mgh=50×9.8×443t=15×60secP=50×9.8×44315×60=241W

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q8. A 75.0-kg painter climbs a ladder that is 2.75 m long leaning against a vertical wall. The ladder makes an angle of

30^{\circ}

angle with the wall. How much work does gravity do on the painter?

A. -2050 J

B. -1750 J

C. -1850 J

D. -1950 J

Answer : Option B
Explaination / Solution:

W=−mghh=xcos30∘=2.75×3–√2W=−75×9.8×2.75×3–√2=−1750J

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q9. A 0.800-kg ball is tied to the end of a string 1.60 m long and swung in a vertical circle. (a) During one complete circle, starting anywhere, calculate the total work done on the ball by (i) the tension in the string and (ii) gravity

A. 30 J, 30 J

B. 25 J, -25 J

C. 0, 0

D. 10 J, 10 J

Answer : Option C
Explaination / Solution:

Work done by conservative force in round trip will be zero (displacement = 0). So that work done by tension and gravity will be zero.

# Work Energy and Power # CBSE 11TH PHYSICS Prepare / Learn

Q10. In which of the following cases is the work done positive?

A. Work done by gravitational force while a man in lifts a bucket out of a well by means of a rope tied to the bucket

B. Work done by friction on a body sliding down an inclined plane

C. work done by an applied force on a body moving on a rough horizontal plane with uniform velocity

D. Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

Answer : Option C
Explaination / Solution:

When a body is moving on a rough horizontal surface then their will be 2 forces acting on the body 1. Applied force ( in the direction of motion) 2. friction ( opposite to direction of motion) As applied force is in same direction as displacement so work done will be positive.

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