Advertisements
Advertisements
प्रश्न
A block is kept on the floor of an elevator at rest. The elevator starts descending with an acceleration of 12 m/s2. Find the displacement of the block during the first 0.2 s after the start. Take g = 10 m/s2.
Advertisements
उत्तर
The free-body diagram of the system is shown below:
The two bodies are separated because the elevator is moving downward with an acceleration of 12 m/s2 (>g) and the body moves with acceleration, g = 10 m/s2 [Freely falling body]
Now, for the block:
g = 10 m/s2, u = 0, t = 0.2 s
So, the distance travelled by the block is given by
\[s = ut + \frac{1}{2}a t^2 \]
\[ = 0 + \frac{1}{2}10 \times \left( 0 . 2 \right)^2 = 5 \times 0 . 04\]
= 0 . 2 m = 20 cm
The displacement of the body is 20 cm during the first 0.2 s.
APPEARS IN
संबंधित प्रश्न
An aircraft executes a horizontal loop at a speed of 720 km/h with its wings banked at 15°. What is the radius of the loop?
Both the springs shown in the following figure are unstretched. If the block is displaced by a distance x and released, what will be the initial acceleration?
A small block B is placed on another block A of mass 5 kg and length 20 cm. Initially, the block B is near the right end of block A (In the following Figure). A constant horizontal force of 10 N is applied to the block A. All the surfaces are assumed frictionless. Find the time that elapses before block B separates from A.
Calculate the tension in the string shown in the following figure. The pulley and the string are light and all the surfaces are frictionless. Take g = 10 m/s2.
Find the mass M of the hanging block in the following figure that will prevent the smaller block from slipping over the triangular block. All the surfaces are frictionless and the strings and the pulleys are light.
Define linear momentum and state its S.I. unit.
Write the mathematical form of Newton's second law of motion. State the conditions if any.
How can Newton's first law of motion be obtained from the second law of motion?
The correct form of Newton's second law is :
A body of mass 5 kg is moving with velocity 2 m s-1. Calculate its linear momentum.
A force acts for 10 s on a stationary body of mass 100 kg, after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate: The velocity acquired by the body.
State the magnitude and direction of the force of gravity acting on the body of mass 5 kg. Take g = 9.8 m s-2.
A ball is thrown vertically upwards. It returns 6 s later. Calculate the greatest height reached by the ball. (Take g = 10 m s−2)
Calculate the velocity of a body of mass 0.5 kg, when it has a linear momentum of 5 Ns.
What do you understand by the term momentum?
What do you mean by an impulsive force?
Use Newton's second law to explain the following:
While catching a fast moving ball, we always pull our hands backwards.
A stone is dropped from a cliff 98 m high.
What will be its speed when it strikes the ground?
A cricket ball of mass 150 g has an initial velocity `u = (3hati + 4hatj)` m s−1 and a final velocity `v = - (3hati + 4hatj)` m s−1 after being hit. The change in momentum (final momentum-initial momentum) is (in kg m s1)
