Work, Energy and Power
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New answer posted
a month agoContributor-Level 10
(1/2)kx² = (1/2)mv² ⇒ x = v√ (m/k) = 10 * √ (4/100) = 10 * (2/10) = 2m
New answer posted
a month agoContributor-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
a month ago
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
a month agoContributor-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
a month agoContributor-Level 9
Deceleration, a = u² / 2S = 10² / (2 * 0.5) = 100 m/s².
Retarding force, F = MA = 0.1 * 100 = 10N
New answer posted
a month agoContributor-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
a month agoContributor-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
a month ago
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
New answer posted
a month agoContributor-Level 10
K? = ½ mv² + ½ Iω²
Since I =? mr² and v=rω,
K? = ½ mv² + ½ (? mr²) (v/r)² = ½ mv² +? mv² = (7/10)mv² = 140 J
K_f = 0.05 K?
(7/10)mv_f² = 0.05 * (140)
(7/10)mv_f² = 7
v_f² = 20
v_f = √20 ≈ 4.47 m/s
New answer posted
a month agoContributor-Level 10
Network done is equal to area under F-x curve
= 200 x 15 + ½ x 100 x 15 + 100 * 15
= 4500 + 750
= 5250
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