Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 m/s Acceleration of the automobile, a = –1.7 ms⁻² From Newton's second law of motion: Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N Hence, the force between the automobile and the road is –2550 N, in the direction opposite toRead more
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 m/s
Acceleration of the automobile, a = –1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N
Hence, the force between the automobile and the road is –2550 N, in the direction
opposite to the motion of the automobile.
(a) 35000 N (b) 1.944 m/s² (c) 15552 N (a) Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F - Ff = 40000 - 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b) Acceleration of the train = a The engine exerts a forcRead more
(a) 35000 N (b) 1.944 m/s² (c) 15552 N
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F – Ff = 40000 – 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass (including the mass of engine), M = m + 8000 = 18000 kg
Fa = Ma
⇒ 𝑎 = 𝐹𝑎/𝑀 = 35000/18000 = 1.944 𝑚/𝑠²
(c) Mass of 4 wagons (excluding wagon 1) m = 2000 × 4 = 8000 kg
Acceleration of the train = 1.944 𝑚/𝑠²
∴ The force of wagon 1 on wagon 2 = 𝑚𝑎 = 8000 × 1.944 = 15552 𝑁
Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 m/s Distance covered by the stone, s = 50 m According to the third equation of motion: v² = u² + 2as Where, Acceleration, a (0)2 = (20)² + 2 × a × 50 a = – 4 m/s² The negative sign indicates that acceleration is acting agaiRead more
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 m/s
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v² = u² + 2as
Where,
Acceleration, a
(0)2 = (20)² + 2 × a × 50
a = – 4 m/s²
The negative sign indicates that acceleration is acting against the motion of the
stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force,
F = Mass x Acceleration
F= ma
F= 1 × (– 4) = – 4 N
Hence, the force of friction between the stone and the ice is – 4 N.
Initial velocity of the truck, u = 0 m/s Time taken, t = 20 s Distance covered by the stone, s = 400 m According to the second equation of motion: 𝑠 = 𝑢𝑡 +1/2 𝑎𝑡² We have 400 = 0 × 20 + 1/2× 𝑎 × (20)² ⇒ 400 = 200𝑎 ⇒ 𝑎 = 2 𝑚/𝑠 ² Now, Force = mass × acceleration ⇒ 𝐹 = 7000 × 2 = 14000 𝑁 For more answeRead more
Initial velocity of the truck, u = 0 m/s
Time taken, t = 20 s
Distance covered by the stone, s = 400 m
According to the second equation of motion:
𝑠 = 𝑢𝑡 +1/2 𝑎𝑡²
We have
400 = 0 × 20 + 1/2× 𝑎 × (20)²
⇒ 400 = 200𝑎
⇒ 𝑎 = 2 𝑚/𝑠 ²
Now, Force = mass × acceleration
⇒ 𝐹 = 7000 × 2 = 14000 𝑁
(c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is reRead more
(c) A batsman hits a cricket ball, which then rolls on a level ground. After
covering a short distance, the ball comes to rest because there is frictional
force on the ball opposing its motion.
Frictional force always acts in the direction opposite to the direction of motion.
Hence, this force is responsible for stopping the cricket ball.
An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 m s–2?
Mass of the automobile vehicle, m = 1500 kg Final velocity, v = 0 m/s Acceleration of the automobile, a = –1.7 ms⁻² From Newton's second law of motion: Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N Hence, the force between the automobile and the road is –2550 N, in the direction opposite toRead more
Mass of the automobile vehicle, m = 1500 kg
Final velocity, v = 0 m/s
Acceleration of the automobile, a = –1.7 ms⁻²
From Newton’s second law of motion:
Force = Mass × Acceleration = 1500 × (–1.7) = –2550 N
Hence, the force between the automobile and the road is –2550 N, in the direction
opposite to the motion of the automobile.
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A 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If the engine exerts a force of 40000 N and the track offers a friction force of 5000 N, then calculate: (a) the net accelerating force; (b) the acceleration of the train; and (c) the force of wagon 1 on wagon 2.
(a) 35000 N (b) 1.944 m/s² (c) 15552 N (a) Force exerted by the engine, F = 40000 N Frictional force offered by the track, Ff = 5000 N Net accelerating force, Fa = F - Ff = 40000 - 5000 = 35000 N Hence, the net accelerating force is 35000 N. (b) Acceleration of the train = a The engine exerts a forcRead more
(a) 35000 N (b) 1.944 m/s² (c) 15552 N
(a) Force exerted by the engine, F = 40000 N
Frictional force offered by the track, Ff = 5000 N
Net accelerating force, Fa = F – Ff = 40000 – 5000 = 35000 N
Hence, the net accelerating force is 35000 N.
(b) Acceleration of the train = a
The engine exerts a force of 40000 N on all the five wagons.
Net accelerating force on the wagons, Fa = 35000 N
Mass of the wagons, m = Mass of a wagon × Number of wagons
Mass of a wagon = 2000 kg
Number of wagons = 5
∴ m = 2000 × 5 = 10000 kg
Total mass (including the mass of engine), M = m + 8000 = 18000 kg
Fa = Ma
⇒ 𝑎 = 𝐹𝑎/𝑀 = 35000/18000 = 1.944 𝑚/𝑠²
(c) Mass of 4 wagons (excluding wagon 1) m = 2000 × 4 = 8000 kg
Acceleration of the train = 1.944 𝑚/𝑠²
∴ The force of wagon 1 on wagon 2 = 𝑚𝑎 = 8000 × 1.944 = 15552 𝑁
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A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Initial velocity of the stone, u = 20 m/s Final velocity of the stone, v = 0 m/s Distance covered by the stone, s = 50 m According to the third equation of motion: v² = u² + 2as Where, Acceleration, a (0)2 = (20)² + 2 × a × 50 a = – 4 m/s² The negative sign indicates that acceleration is acting agaiRead more
Initial velocity of the stone, u = 20 m/s
Final velocity of the stone, v = 0 m/s
Distance covered by the stone, s = 50 m
According to the third equation of motion:
v² = u² + 2as
Where,
Acceleration, a
(0)2 = (20)² + 2 × a × 50
a = – 4 m/s²
The negative sign indicates that acceleration is acting against the motion of the
stone.
Mass of the stone, m = 1 kg
From Newton’s second law of motion:
Force,
F = Mass x Acceleration
F= ma
F= 1 × (– 4) = – 4 N
Hence, the force of friction between the stone and the ice is – 4 N.
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A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 metric tonnes (Hint: 1 metric tonne = 1000 kg.)
Initial velocity of the truck, u = 0 m/s Time taken, t = 20 s Distance covered by the stone, s = 400 m According to the second equation of motion: 𝑠 = 𝑢𝑡 +1/2 𝑎𝑡² We have 400 = 0 × 20 + 1/2× 𝑎 × (20)² ⇒ 400 = 200𝑎 ⇒ 𝑎 = 2 𝑚/𝑠 ² Now, Force = mass × acceleration ⇒ 𝐹 = 7000 × 2 = 14000 𝑁 For more answeRead more
Initial velocity of the truck, u = 0 m/s
Time taken, t = 20 s
Distance covered by the stone, s = 400 m
According to the second equation of motion:
𝑠 = 𝑢𝑡 +1/2 𝑎𝑡²
We have
400 = 0 × 20 + 1/2× 𝑎 × (20)²
⇒ 400 = 200𝑎
⇒ 𝑎 = 2 𝑚/𝑠 ²
Now, Force = mass × acceleration
⇒ 𝐹 = 7000 × 2 = 14000 𝑁
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A batsman hits a cricket ball which then rolls on a level ground. After covering a short distance, the ball comes to rest. The ball slows to a stop because
(c) A batsman hits a cricket ball, which then rolls on a level ground. After covering a short distance, the ball comes to rest because there is frictional force on the ball opposing its motion. Frictional force always acts in the direction opposite to the direction of motion. Hence, this force is reRead more
(c) A batsman hits a cricket ball, which then rolls on a level ground. After
covering a short distance, the ball comes to rest because there is frictional
force on the ball opposing its motion.
Frictional force always acts in the direction opposite to the direction of motion.
Hence, this force is responsible for stopping the cricket ball.
For more answers visit to website:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/