Work, Energy and Power

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New answer posted

2 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

T = W + f

= 20000 + 3000

= 23000 N

Power = T v

= 23000 * 1.5

= 34500 watts 

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

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.

New answer posted

2 months ago

0 Follower 5 Views

V
Vishal Baghel

Contributor-Level 10

(1/2)kx² = (1/2)mv² ⇒ x = v√ (m/k) = 10 * √ (4/100) = 10 * (2/10) = 2m

New answer posted

2 months ago

0 Follower 3 Views

V
Vishal Baghel

Contributor-Level 10

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

New answer posted

2 months ago

0 Follower 6 Views

V
Vishal Baghel

Contributor-Level 10

(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 differentiation

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

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

New answer posted

2 months ago

0 Follower 3 Views

R
Raj Pandey

Contributor-Level 9

Deceleration, a = u² / 2S = 10² / (2 * 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 * 100 = 10N

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

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

New answer posted

2 months ago

0 Follower 19 Views

A
alok kumar singh

Contributor-Level 10

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 = √* (gh/2)* * (1 + e²) / (1 + e)² = 5 * (1.81) / (1.9)² = 2.50 m/s

New answer posted

2 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

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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