1.A lift weighting 1000kg is moving upwards with an acceleration of 1m/s^2.The tension in the supporting cable is
(a)980N (b)10800N (c)9800N (d)8800N
2.A person of mass 60kg is inside a lift of mass 940kg and presses the button on control panel.The starts moving upwards with an acceleration 1.0m/s^2.If g=10m/s^2,the tension in the supporting cable is
(a)8600N (b)9680N (c)11000N (d)1200N
3.A rubber ball is dropped from a height of 5 metre on a plane,where the acceleration due to gravity is not shown.On bouncing it rises to 1.8metre.The ball loses its velocity on bouncing by a factor of
(a)16/25(b)2/5(c)3/5(d)9/25
1000*9.81 + 1000*1 = 10810 N
total mass = 1000 kg
F = 1000*10 + 1000 * 1 = 11,000 N
Ke on collision = m g h = m g (5)
so
(1/2) m v^2 = 5 m g
v^2 = 10 g
on rebound v^2 = 3.6 g
sqrt (3.6/10) = sqrt .36 = sqrt (36/100)
= 6/10 = 3/5
To solve these questions, we can use the concepts of Newton's second law and the principle of conservation of energy.
1. To find the tension in the supporting cable of the lift, we need to consider the net force acting on the lift. The net force can be determined using Newton's second law, which states that force (F) is equal to mass (m) multiplied by acceleration (a).
Given:
Mass of lift (m) = 1000 kg
Acceleration (a) = 1 m/s^2
Using the formula F = ma, we can calculate the net force acting on the lift:
F = 1000 kg * 1 m/s^2
F = 1000 N
Since the net force acting on the lift is equal to the tension in the supporting cable, the tension is 1000 N. Therefore, the answer is (a) 980 N.
2. To find the tension in the supporting cable of the lift in this scenario, we need to consider the forces acting on the person and the lift. The total force on the person is their weight (mg), where m is the mass and g is the acceleration due to gravity.
Given:
Mass of the person (m) = 60 kg
Mass of the lift (M) = 940 kg
Acceleration (a) = 1.0 m/s^2
Acceleration due to gravity (g) = 10 m/s^2
The net force acting on the person is:
F = mg
F = 60 kg * 10 m/s^2
F = 600 N
The net force acting on the lift is:
F = Ma
F = 940 kg * 1.0 m/s^2
F = 940 N
The tension in the supporting cable is the sum of the forces acting on the person and the lift:
Tension = (mg) + (Ma)
Tension = 600 N + 940 N
Tension = 1540 N
Therefore, the answer is (none of the provided options). None of the options match the calculated tension of 1540 N.
3. To determine the factor by which the velocity of the ball changes upon bouncing, we can use the principle of conservation of energy. The potential energy of the ball before bouncing is converted into kinetic energy after bouncing.
Given:
Initial height (h1) = 5 m
Final height (h2) = 1.8 m
The potential energy before bouncing is equal to the kinetic energy after bouncing:
mgh1 = 1/2 mv2^2
Since the mass of the ball cancels out, we can write:
gh1 = 1/2 v2^2
Dividing both sides of the equation by gh1, we get:
1/2 v2^2 = h1/h2
For the factor by which the velocity changes, we can take the square root of both sides of the equation:
v2/v1 = √(h1/h2)
Substituting the given values, we have:
v2/v1 = √(5/1.8)
v2/v1 ≈ √2.7778
v2/v1 ≈ 1.6667
Therefore, the ball loses its velocity on bouncing by a factor of approximately 1.6667. None of the provided options match this factor.