Chapter 7 – Motion, Force and Work
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 ________.
Ans: constant
(b) If a body is moving with a constant velocity its acceleration is ________.
Ans: zero
(c) ________ is a scalar quantity.
Ans: Speed
(d) ________ is the distance traversed by a body in a particular direction in unit time.
Ans: Velocity
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?
Ans:
The distance covered by Sachin and Sameer AB (3 km), BC (4 km), CD (5 km), DE (3 km)
Total distance = 3 + 4 + 5 + 3
∴ Total distance = 15 km
The actual distance covered is 15 km.
Total displacement from A → E = 3 + 3 + 3
∴ Total displacement from A = 9 km
Speed = \(\large \frac {Distance}{Time}\)
∴ Speed = \(\large \frac {15\,km}{1\, hour}\)
∴ Speed = 15 km/hour
Velocity = \(\large \frac {Displacement}{Time}\)
∴ Velocity = \(\large \frac {9\,km}{1\,hour}\)
∴ Velocity = 9 km/hr
Ans: The velocity from A to E is 15 km/hr which can be average velocity.
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 |
Ans:
| A | B | C |
|---|---|---|
Work | Joule | erg |
Force | Newton | dyne |
Displacement | Metre | cm |
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.
Ans:
(i) The bird takes a circular turn and comes back.
(ii) The distance measured by the bird equals the circular path it has taken.
(iii) Since the bird comes back to its original position, the displacement will be zero.
5. 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 lives by using force. For example, Lifting luggage, pushing a vehicle, riding a bicycle, or taking things from lower floors to upper ones The work done by force (W = F × S) Here, the displacement is taking place in the direction of force.
(i) When a handcart is pushed with a certain force, some work is done.
(ii) Lifting our school bag and carrying it to school is also work.
(iii) If we cover some distance and go to school from home, displacement takes place.
(iv) We keep changing velocity during this travel.
(v) When we travel by vehicle, the distance covered at a certain velocity is not constant.
(vi) If the velocity is increased, the acceleration becomes positive, whereas when the brake is applied to the vehicle, the acceleration becomes negative.
6. A ball is rolling from A to D on a flat and smooth surface. Its speed is 2 cm/s. On reaching B, it was pushed continuously up to C. On reaching D from C, its speed had become 4 cm/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:
(i) The ball is rolling on a flat and smooth surface. Therefore, it will not face the frictional resistance.
Therefore, the speed of ball from A to B is 2 cm/sec
(ii) At B the speed of the ball is 2 cm/s. From C to D there is a force acting on a ball. Therefore, the speed of the ball at D is 4 cm/s.
(iii) The speed of the ball is the same at all the points as it is in linear motion. Therefore, the magnitude of velocity of ball = speed of the ball.
(iv) Therefore, increase in the speed from B to C,
= 4 cm/s – 2 cm/s
= 2 cm/s.
(v) Acceleration in the displacement
= \(\large \frac {Change\,in\, velocity}{time}\)
= \(\large \frac {2\,cm/s}{2\,s}\)
= 1 cm/s²
(vi) Thus, the acceleration between B to C is 1 cm/s².
7. Solve the following problems.
(a) 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 = 1000N
s = – 10m
We know that,
W = F × s
∴ W = 1000N (– 10m)
∴ W = – 10000 J
Ans: The work done is – 10000 J.
(b) 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:
F = 2N
s = 50m
We know that,
W = F × s
∴ W = 2N × 50m
∴ W = 100 J
Ans: The work done by the force is – 100J.