7- Motion, Force and Work
1.
Fill
in the blanks with the proper words from the brackets :
Ans (1) constant (2) zero (3) speed (4)velocity
Q 2. Explain the following concepts in your own
words with everyday examples:
force
work, displacement, velocity, acceleration, distance.
Ans.
We perform many activities in our daily life by suing force. e.g. lifting the
luggage. pushing a vehicle, riding a bicycle, taking things from lower floor to
upper ones. The work done by force (F), W = F x s. Here s is the displacement
taking place in the direction of the force.
When a hand cart is pushed with a
certain force some work is done. Lifting our school bag and carrying it to
school is also a work. We cover some distance as we go to school from home the
displacement takes place. We keep on changing velocity during this travel. When
we travel by vehicle, the distance is crossed with a certain velocity. But this
velocity is not constant. If the velocity is increased, the acceleration
becomes positive whereas when the brake is applied on the vehicle, the
acceleration becomes negative.
Q. 3.
A ball is rolling from A to D on a flat and smooth surface. Its speed is 2cm/s.
On reaching B, it was pushed continuously up to C. On reaching D from C, its
speed had become 4cm/s. It took 2 seconds for it to go from B to C. What is the
acceleration of the ball as it goes from B to C?
Ans.
The ball moves as follows:
The
speed of ball from A to B is 2 cm/sec.
At B
the speed of ball is 2 cm/second.
From C
to D there is force acting on a ball. Therefore, the speed of the ball is 4
cm/sec (the speed of the ball at D)
The
speed of the ball is same at all the points as it is linear motion.Therefore,
the magnitude of velocity of ball = speed of the ball.
Therefore,
increase in the speed from B to C = 4 cm/sec – 2 cm/sec = 2 cm/sec
Acceleration
in the displacement = change in the velocity/ time = 2 cm/s/2 sec = 1 cm/s2
Thus
the acceleration between Bto C = 1cm/s2
Q 4. Observe
the figure and answer the questions:
Ans. The distance covered by Sachin and Sameer
:
A → B
(3km), B → C (4 km), C → D (5km), D → E (3 km)
Total
distance : 3 + 4 + 5 + 3 = 15 km
The
actual distance covered = 15 km.
Total
displacement: From A to E = 3 + 3 + 3 = 9 km
Total
displacement = 9 km
Speed
= Distance/ Time =15 km/1 hour =15 km/hour
Velocity
= Displacement /Time = 9km/1 hour = 9 km/hour
The
velocity from A to E is 15 km/hour which can be called the average velocity
Q. 5. From the groups B and C, choose the proper
words, for each of the words in
group A:
Ans
Work Joule Erg
Force Newton dyne
Displacement Metre
cm
Q. 6. 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.
Ans.
The bird takes a circular turn and comes back. The distance traversed by the
bird is equal to the circular path that it has taken. Since bird comes back to
its original position the displacement will be zero.
(Note:
The path of a bird need not be in one plant. It usually flies in all the three
planes. While deciding the displacement, we take into account.)
Q 7.
Solve the following problems :
(1) A
force of 1000N was applied to stop a car that was moving with a constant velocity.
The car stopped after moving through 10m. How much is the work done?
Solution
: Here the direction of force and displacement are opposite to each other.
That
means
F =
1000 N and
s = -
10 m
. W = F x s
= 1000 NX (-10 m)
= - 10000 J
Ans. The work done W = -10000J
(2) A
cart with mass 20 kg went 50m in a straight line on a plain and smooth road
when a force of 2N was applied to it. How much work was done by the force?
Solution
: Force (F) = 2 N
Displacement (s) = 50 metre
W = F x s
(work done by force)
W = 2 N X 50 m
= 100J
Ans.
The work done by the force W = -100J
