Std.VII- Motion,Force and Work (Question&Answers)

Question 1:

Fill in the blanks with the proper words from the brackets.
(stationary, zero, changing, constant, displacement, velocity, speed, acceleration, stationary but not zero, increases)
(a) If a body traverses a distance in direct proportion to the time, the speed of the body is ............... .
(b) If a body is moving with a constant velocity its acceleration is .......... .
(c) .............. is a scalar quantity.
(d) ............. is the distance traversed by a body in a particular direction in unit time.

ANSWER:

(a) If a body traverses a distance in direct proportion to the time, the speed of the body is constant.
(b) If a body is moving with a constant velocity its acceleration is zero.
(c) Speed is a scalar quantity.
(d) Velocity is the distance traversed by a body in a particular direction in unit time.

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Question 2:

Observe the figure and answer the questions.

Sachin and Sameer started on a motorbike from place A, took the turn at B, did a task at C, travelled by the route CD to D and then went on to E. Altogether, they took one hour for this journey. Find out the actual distance traversed by them and the displacement from A to E. From this, deduce their speed. What was their velocity from A to E in the direction AE? Can this velocity be called average velocity?

ANSWER:

.

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Question 3:

From the groups B and C, choose the proper words, for each of the words in group A.
 

A	B	C
Work	Newton	erg
Force	Metre	cm
Displacement	Joule	dyne

ANSWER:

A	B	C
Work	Joule	erg
Force	Newton	dyne
Displacement	Metre	cm

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Question 4:

A bird sitting on a wire, flies, circles around and comes back to its perch. Explain the total distance it traversed during its flight and its eventual displacement.

ANSWER:

The total distance travelled by the bird during its flight = 2×$\times$(Distance between the point where the bird was sitting and the point from where it takes a turn)
The eventual displacement of the bird is zero as it returns to its initial point i.e. where it was sitting.

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Question 5:

Explain the following concepts in your own words with everyday examples : force, work, displacement, velocity, acceleration, distance.

ANSWER:

Force: It is any kind of push or pull on a body due to another body when the bodies interact with each other. It is a vector quantity. For example, a person applies a force in the form of push or pull to open a door.

Work: It is defined as the work done by a force that causes a displacement in an object. It is a scalar quantity. For example, a child does work when he drags a toy car on the ground.

Displacement: It is the shortest distance between the initial and final position of an object. It is a vector quantity. For example, when we go to a mall for shopping from our house and then return to the house, the displacement would be 0 as our initial position (house) and final position (house) is same.

Velocity: Velocity is defined as the rate of change of displacement. It is a vector quantity. For example, a car running on a straight road has some velocity.

Acceleration: The rate of change of velocity is known as acceleration. It is a vector quantity. For example, if a car is moving on a straight with variable speed, it will posses some acceleration. In case the speed of the car remains same bu

 the direction of car changes, then also the car would be accelerating.

Distance: It is the actual path length covered by an object during its motion. It is a scalar quantity.  For example, when we go to a mall for shopping from our house and then return to the house, the distance travelled by us would be twice the distance between our house and the mall.

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Question 7:

Solve the following problems.
(a) A force of 1000 N was applied to stop a car that was moving with a constant velocity. The car stopped after moving through 10 m. How much is the work done?
(b) A cart with mass 20 kg went 50 m in a straight line on a plain and smooth road when a force of 2 N was applied to it. How much work was done by the force?

ANSWER:

(a) Work done by the force to stop the car=

F×S=1000×10=10000 J $\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the}$

$\mathrm{force} \mathrm{to} \mathrm{stop} \mathrm{the} \mathrm{car}=F\times S=1000\times 10=10000 \mathrm{J}$

(b) Work done by the force=F×S=2×50=100 J$\mathrm{Work} \mathrm{done} \mathrm{by} \mathrm{the} \mathrm{force}=F\times S=2\times 50=$100 J