When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ when there is friction, the downward acceleration of the block is a´ = g (sin θ – μ cos θ) As the block slides a distance d in each case so d= 1/2 at² = 1/2 a't'² a/a' = t'²/t ² = (nt)²/t² = n² or gsinRead more
When there is no friction, the block slides down the inclined plane with
acceleration.
a = g sin θ
when there is friction, the downward acceleration of the block is
a´ = g (sin θ – μ cos θ)
As the block slides a distance d in each case so
d= 1/2 at² = 1/2 a’t’²
A bob of mass 0.1 kg hung from the ceiling of room by a string 2 m long is oscillating. At its mean position the speed of a bob is 1 ms–1. What is the trajectory of the oscillating bob if the string is cut when the bob is (i) At the mean position (ii) At its extreme position
(i) Parabolic, (ii) vertically downwards
(i) Parabolic, (ii) vertically downwards
See lessA spring balance is attached to the ceiling of a lift. When the lift is at rest spring balance reads 49 N of a body hang on it. If the lift moves : (i) Downward (ii) upward, with an acceleration of 5 ms⁻² (iii) with a constant velocity. What will be the reading of the balance in each case?
(i) R = m (g – a) weight = 49 N so m = 49/9.8 = 5kg R = 5 (9·8 – 5) R = 24N (ii) R = m (g + a) R = 5 (9·8 + 5) R = 74 N (iii) as a = 0 so R = mg = 49N
(i) R = m (g – a)
weight = 49 N
so m = 49/9.8 = 5kg
R = 5 (9·8 – 5)
R = 24N
(ii) R = m (g + a)
R = 5 (9·8 + 5)
R = 74 N
(iii) as a = 0 so R = mg = 49N
See lessSmooth block is released at rest on a 45° incline and then slides a distance d. If the time taken of slide on rough incline is n times as large as that to slide than on a smooth incline. Show that coefficient of friction. µ = (1-1/n ²).
When there is no friction, the block slides down the inclined plane with acceleration. a = g sin θ when there is friction, the downward acceleration of the block is a´ = g (sin θ – μ cos θ) As the block slides a distance d in each case so d= 1/2 at² = 1/2 a't'² a/a' = t'²/t ² = (nt)²/t² = n² or gsinRead more
When there is no friction, the block slides down the inclined plane with
acceleration.
a = g sin θ
when there is friction, the downward acceleration of the block is
a´ = g (sin θ – μ cos θ)
As the block slides a distance d in each case so
d= 1/2 at² = 1/2 a’t’²
a/a’ = t’²/t ² = (nt)²/t² = n²
See lessor gsinθ/g(sinθ – μ cosθ) = n²
A 50 g bullet is fired from a 10 kg gun with a speed of 500 ms⁻¹. What is the speed of the recoil of the gun?
Initial momentum = 0 using conservation of linear momentum mv + MV = 0 V= -mv/m ⇒ v = 2.5 m/s
Initial momentum = 0
See lessusing conservation of linear momentum
mv + MV = 0
V= -mv/m
⇒ v = 2.5 m/s
A particle of mass 0.3 kg is subjected to a force of F = – kx with k = 15 Nm⁻¹ . What will be its initial acceleration if it is released from a point 20 cm away from the origin?
As F = ma so F = – kx = ma a= -kx/m for x = 20 cm, ⇒ a = – 10 m/s²
As F = ma so F = – kx = ma
See lessa= -kx/m
for x = 20 cm, ⇒ a = – 10 m/s²