Work, Energy and Power
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The potential energy of a long spring when stretched by is . If the spring is stretched by , potential energy stored in it will be:
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The potential energy of a long spring when stretched by is . If the spring is stretched by , potential energy stored in it will be:
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The angular acceleration direction is given along angular velocity or opposite to angular velocity depending upon whether angular velocity magnitude is increasing or decreasing and this direction remains along the axis of circular motion.
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A ball of mass 4 kg, moving with a velocity 10 ms?¹, collides with a spring of length 8m and force constant 100 Nm?¹. The length of compressed spring is x m. The value of x, to the nearest integer, is.
A ball of mass 4 kg, moving with a velocity 10 ms?¹, collides with a spring of length 8m and force constant 100 Nm?¹. The length of compressed spring is x m. The value of x, to the nearest integer, is.
(1/2)kx² = (1/2)mv² ⇒ x = v√ (m/k) = 10 × √ (4/100) = 10 × (2/10) = 2m
Upvote An object of mass m₁ collides with another object of mass m₂, which is at rest. After collision the objects move with equal speeds in the opposite direction. The ratio of the masses m₂: m₁ is :
An object of mass m₁ collides with another object of mass m₂, which is at rest. After collision the objects move with equal speeds in the opposite direction. The ratio of the masses m₂: m₁ is :
With the help of definition of e, we can write
e = v? /v? = 2v/u ⇒ u = 2v/e . (2)
Putting the value of e in equation (1), we have
m? (2v/e) = (m? - m? )v ⇒ 2m? = em? - em? ⇒ m? /m? = (2+e)/e = 1 + 2/e > 2
Upvote The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by :
The velocity-displacement graph of a particle is shown in the figure. The acceleration-displacement graph of the same particle is represented by :
(v/v? ) + (x/x? ) = 1 ⇒ v = - (v? /x? )x + v?
⇒ a = dv/dt = - (v? /x? ) (v) = - (v? /x? ) [- (v? /x? )x + v? ] ⇒ a = (v? ²/x? ²)x - v? ²/x?
- Concept involved: Graph of kinematics
- Topic: Kinematics
- Difficulty level: Moderate
- Note: IIT-Jee-2005
- Point of Error: Writing Equation of straight line and differe
Upvote As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s²) The value of 'x' to the nearest integer is----------.
As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed of the particle at B is x m/s. (Take g = 10 m/s²) The value of 'x' to the nearest integer is----------.
Using Conservation of Mechanical Energy at point-A and at point-B, we can write
K_B = U_A - U_B [Since K_A = 0]
⇒ (1/2)mv_B² = mg (h_A - h_B)
⇒ v_B = √ (2 × 10 × (10 - 5) = 10m/s
Upvote A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10m/s and slows down with uniform deceleration, then magnitude effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is----------.
A bullet of mass 0.1 kg is fired on a wooden block to pierce through it, but it stops after moving a distance of 50 cm into it. If the velocity of bullet before hitting the wood is 10m/s and slows down with uniform deceleration, then magnitude effective retarding force on the bullet is 'x' N. The value of 'x' to the nearest integer is----------.
Deceleration, a = u² / 2S = 10² / (2 × 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 × 100 = 10N
Upvote A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to:
A constant power delivering machine has towed a box, which was initially at rest, along a horizontal straight line. The distance moved by the box in time 't' is proportional to:
m (dv/dt) = P ⇒ ∫v dv = ∫ (P/m) dt ⇒ v = (2Pt/m)¹/²
⇒ ∫dx = ∫ (2P/m)¹/² t¹/² dt ⇒ x = (2P/m)¹/² (2/3)t³/² ⇒ x ∝ t³/²
Upvote A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81/100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms⁻²)
A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to 81/100 of the height through which it falls. Find the average speed of the ball. (Take g = 10 ms⁻²)
For a bouncing object with initial height h = 5m and coefficient of restitution e = 0.9 (so e² =0.81):
Total distance traveled, d = h + 2e²h + 2e? h + . = h * (1 + e²) / (1 - e²)
Total time taken, t = √* (2h/g)* + 2√* (2e²h/g)* + 2√* (2e? h/g)* + . = √* (2h/g)* * (1 + e) / (1 - e)
Average speed = d/t =
Upvote A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms?¹ goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?
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A sphere of mass 2 kg and radius 0.5 m is rolling with an initial speed of 1 ms?¹ goes up an inclined plane which makes an angle of 30° with the horizontal plane, without slipping. How long will the sphere take to return to the starting point A?
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The acceleration 'a' is calculated as:
a = (g sinθ) / (1 + I/mr²) = (10 * sin30°) / (1 + 2/5) = 25/7 m/s²
The time 't' is calculated as:
t = 2v / a = (2 * 1) / (25/7) = 0.57 s
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