We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula. p = mv, whereas K.E. = (1/2) mv². Express the Kinetic Energy as a Function of Momentum: Let's convert Kinetic energy as a function of momeRead more
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula.
p = mv, whereas K.E. = (1/2) mv².
Express the Kinetic Energy as a Function of Momentum:
Let’s convert Kinetic energy as a function of momentum; since momentum was multiplied by constant so was velocity by same amount:
K.E. = (1/2) mv = (1/2)(p/v).
From p = mv, we have:
v = p/m
Substituting v in the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Calculate the initial and final momentum
Let the initial momentum be p. If momentum increases by 20%, the new momentum (p’) is:
p’ = p + 0.2p = 1.2p
Step 4: Calculate initial and final kinetic energy
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.). Step 1: Write the formulas - Momentum (p) is given by: p = mv - Kinetic energy (K.E.) is given by: K.E. = (1/2) mv² Step 2: Relate momentum to kiRead more
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.).
Step 1: Write the formulas
– Momentum (p) is given by:
p = mv
– Kinetic energy (K.E.) is given by:
K.E. = (1/2) mv²
Step 2: Relate momentum to kinetic energy
We can write kinetic energy in terms of momentum:
Using the relation p = mv, we can rearrange to find v:
v = p/m
Substituting this into the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Substitute the values
Given:
– Mass (m) = 2 kg
– Momentum (p) = 2 N·s (Note: momentum is in kg·m/s which is equivalent to N·s)
Now substituting into the formula for kinetic energy:
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force. Given: Mass of the body (m) = 2 kg Coefficient of friction (μ) = 0.2 Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity) Step 1: Calculate theRead more
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force.
Given:
Mass of the body (m) = 2 kg
Coefficient of friction (μ) = 0.2
Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity)
Step 1: Calculate the normal force (N)
On a horizontal plane, the normal force (N) is equal to the weight of the body:
N = m * g
N = 2 kg * 10 m/s²
N = 20 N
Step 2: Find the maximum static friction (f_s)
The maximum static friction is given by:
f_s = μ * N
f_s = 0.2 * 20 N
f_s = 4 N
Step 3: Analyze the applied force (F)
Case (a): If F = 5 N
– The applied force (5 N) is greater than the maximum static friction (4 N).
– Hence, the body will move in the forward direction.
Case (b): If F = 3 N
– The applied force (3 N) is less than the maximum static friction (4 N).
– Thus, the body will not move and will remain at rest.
**Case (c):** If F = 3 N
– Since the applied force is not enough to overcome friction, the body will remain at rest.
Conclusion:
– For F = 5 N, the body will move in the forward direction.
– For F = 3 N, the body will remain at rest.
– Hence, both (a) and (c) are correct.
Final Answer:
The correct options are both (a) and (c).
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion. 1. Work done will be zero: - True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body. 2. Acceleration will be zeroRead more
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion.
1. Work done will be zero:
– True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body.
2. Acceleration will be zero:
– False. The velocity has a constant speed but is changing direction all the time, and hence there is a centripetal acceleration directed toward the center of the circular path.
3. No force acts on the body:
– False. A net force (centripetal force) is necessary to keep the body moving in a circular path and is directed towards the center.
4. Its velocity remains constant:
– False. True, the velocity magnitude is unchanged, but the velocity is a vector: its direction must change; it is not a constant.
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension of the spring Step 1: ReRead more
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension of the spring
Step 1: Relationship between force and extension
According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:
F = kx
Rearranging gives us:
x = F/k
Step 2: Calculate potential energy for both springs
For spring 1 (k₁ = 1500 N/m):
U₁ = (1/2) k₁ x₁²
Substituting for x₁:
U₁ = (1/2) k₁ (F/k₁)²
U₁ = (1/2) k₁ (F²/k₁²)
U₁ = (1/2) (F²/k₁)
For spring 2 (k₂ = 3000 N/m):
U₂ = (1/2) k₂ x₂²
Substituting for x₂:
U₂ = (1/2) k₂ (F/k₂)²
U₂ = (1/2) k₂ (F²/k₂²)
U₂ = (1/2) (F²/k₂)
If momentum is increased by 20 %, then kinetic energy increases by
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula. p = mv, whereas K.E. = (1/2) mv². Express the Kinetic Energy as a Function of Momentum: Let's convert Kinetic energy as a function of momeRead more
We can then solve for how kinetic energy would have increased for that increase of momentum by 20%. So first we give out the momentum and kinetic energy formula.
p = mv, whereas K.E. = (1/2) mv².
Express the Kinetic Energy as a Function of Momentum:
Let’s convert Kinetic energy as a function of momentum; since momentum was multiplied by constant so was velocity by same amount:
K.E. = (1/2) mv = (1/2)(p/v).
From p = mv, we have:
v = p/m
Substituting v in the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Calculate the initial and final momentum
Let the initial momentum be p. If momentum increases by 20%, the new momentum (p’) is:
p’ = p + 0.2p = 1.2p
Step 4: Calculate initial and final kinetic energy
Initial K.E. (K.E.₁):
K.E.₁ = (1/2) (p²/m)
Final K.E. (K.E.₂):
K.E.₂ = (1/2) (p’²/m)
K.E.₂ = (1/2) [(1.2p)²/m]
K.E.₂ = (1/2) [1.44p²/m]
Step 5: Find the increase in kinetic energy
Now, we calculate the increase in kinetic energy:
Increase in K.E. = K.E.₂ – K.E.₁
Increase in K.E. = (1/2) [1.44p²/m] – (1/2) [p²/m]
Increase in K.E. = (1/2m) [1.44p² – p²]
Increase in K.E. = (1/2m) [0.44p²]
Percentage increase in K.E. is given by:
Percentage increase = (Increase in K.E. / K.E.₁) × 100
Percentage increase = [(0.44p²/2m) / (p²/2m)] × 100
Percentage increase = 0.44 × 100
Percentage increase = 44%
Final Answer:
See lessIf the momentum is increased by 20%, then it increases the Kinetic energy with 44%.
The kinetic energy of body of mass 2 kg and momentum of 2 N is
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.). Step 1: Write the formulas - Momentum (p) is given by: p = mv - Kinetic energy (K.E.) is given by: K.E. = (1/2) mv² Step 2: Relate momentum to kiRead more
To calculate the kinetic energy of a body given its mass and momentum, we can make use of the relationship between momentum (p) and kinetic energy (K.E.).
Step 1: Write the formulas
– Momentum (p) is given by:
p = mv
– Kinetic energy (K.E.) is given by:
K.E. = (1/2) mv²
Step 2: Relate momentum to kinetic energy
We can write kinetic energy in terms of momentum:
Using the relation p = mv, we can rearrange to find v:
v = p/m
Substituting this into the kinetic energy formula:
K.E. = (1/2) m (p/m)²
K.E. = (1/2) (p²/m)
Step 3: Substitute the values
Given:
– Mass (m) = 2 kg
– Momentum (p) = 2 N·s (Note: momentum is in kg·m/s which is equivalent to N·s)
Now substituting into the formula for kinetic energy:
K.E. = (1/2) (p²/m)
K.E. = (1/2) [(2)² / 2]
K.E. = (1/2) [4 / 2]
K.E. = (1/2) [2]
K.E. = 1 J
Final Answer:
The kinetic energy of the body is 1 J.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
A body of mass 2 kg is placed on rough horizontal plane. The coefficient of friction between body and plane is 0.2 Then
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force. Given: Mass of the body (m) = 2 kg Coefficient of friction (μ) = 0.2 Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity) Step 1: Calculate theRead more
To find the motion of the body placed on a rough horizontal plane, we need to consider the effects of friction and applied force.
Given:
Mass of the body (m) = 2 kg
Coefficient of friction (μ) = 0.2
Acceleration due to gravity (g) = 9.8 m/s² (we will use 10 m/s² for simplicity)
Step 1: Calculate the normal force (N)
On a horizontal plane, the normal force (N) is equal to the weight of the body:
N = m * g
N = 2 kg * 10 m/s²
N = 20 N
Step 2: Find the maximum static friction (f_s)
The maximum static friction is given by:
f_s = μ * N
f_s = 0.2 * 20 N
f_s = 4 N
Step 3: Analyze the applied force (F)
Case (a): If F = 5 N
– The applied force (5 N) is greater than the maximum static friction (4 N).
– Hence, the body will move in the forward direction.
Case (b): If F = 3 N
– The applied force (3 N) is less than the maximum static friction (4 N).
– Thus, the body will not move and will remain at rest.
**Case (c):** If F = 3 N
– Since the applied force is not enough to overcome friction, the body will remain at rest.
Conclusion:
– For F = 5 N, the body will move in the forward direction.
– For F = 3 N, the body will remain at rest.
– Hence, both (a) and (c) are correct.
Final Answer:
The correct options are both (a) and (c).
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
When a body moves with constant speed in a circular path, then
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion. 1. Work done will be zero: - True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body. 2. Acceleration will be zeroRead more
When a body is moving in a circular path with constant speed, we have to look into the consequences of circular motion.
1. Work done will be zero:
– True. Since the force (centripetal force) is always perpendicular to the direction of motion, no work is done on the body.
2. Acceleration will be zero:
– False. The velocity has a constant speed but is changing direction all the time, and hence there is a centripetal acceleration directed toward the center of the circular path.
3. No force acts on the body:
– False. A net force (centripetal force) is necessary to keep the body moving in a circular path and is directed towards the center.
4. Its velocity remains constant:
– False. True, the velocity magnitude is unchanged, but the velocity is a vector: its direction must change; it is not a constant.
Final Answer:
Work done will be zero.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/
Two springs of spring constant 1500 N/m and 3000 N/m respectively are stretched with a same force. Their potential energies will be in the ratio of
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy: U = (1/2) k x² Where: - U is the potential energy - k is the spring constant - x is the extension of the spring Step 1: ReRead more
To find the ratio of the potential energies of two springs with different spring constants when stretched by the same force, we can use the formula for elastic potential energy:
U = (1/2) k x²
Where:
– U is the potential energy
– k is the spring constant
– x is the extension of the spring
Step 1: Relationship between force and extension
According to Hooke’s Law, the force applied to a spring is related to the spring constant and the extension by:
F = kx
Rearranging gives us:
x = F/k
Step 2: Calculate potential energy for both springs
For spring 1 (k₁ = 1500 N/m):
U₁ = (1/2) k₁ x₁²
Substituting for x₁:
U₁ = (1/2) k₁ (F/k₁)²
U₁ = (1/2) k₁ (F²/k₁²)
U₁ = (1/2) (F²/k₁)
For spring 2 (k₂ = 3000 N/m):
U₂ = (1/2) k₂ x₂²
Substituting for x₂:
U₂ = (1/2) k₂ (F/k₂)²
U₂ = (1/2) k₂ (F²/k₂²)
U₂ = (1/2) (F²/k₂)
Step 3: Find the ratio of potential energies
Now we can find the ratio of U₁ to U₂:
U₁ / U₂ = (F² / (2k₁)) / (F² / (2k₂))
U₁ / U₂ = k₂ / k₁
U₁ / U₂ = 3000 / 1500
U₁ / U₂ = 2
Final Answer:
The potential energies of the two springs will be in the ratio of 2 : 1.
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See lesshttps://www.tiwariacademy.com/ncert-solutions/class-11/physics/chapter-5/