English
Karnataka Board PUCPUC Science Class 11

Two Blocks of Equal Mass M Are Tied to Each Other Through a Light String. One of the Blocks is Pulled Along the Line Joining Them with a Constant Force F.

Advertisements
Advertisements

Question

Two blocks of equal mass m are tied to each other through a light string. One of the blocks is pulled along the line joining them with a constant force F. Find the tension in the string joining the blocks.

Sum
Advertisements

Solution



Let a be the common acceleration of the blocks.
For block 1,
\[F - T = ma\]
For block 2,
T = ma    ...(2)
Subtracting equation (2) from (1), we get:
\[F - 2T = 0\]
\[\Rightarrow T = \frac{F}{2}\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 5: Newton's Laws of Motion - Exercise [Page 79]

APPEARS IN

HC Verma Concepts of Physics Volume 1 and 2 [English]
Chapter 5 Newton's Laws of Motion
Exercise | Q 5 | Page 79

RELATED QUESTIONS

The below figure shows the position-time graph of a particle of mass 4 kg.

  1. What is the force on the particle for t < 0, t > 4 s, 0 < t < 4 s?
  2. What is the impulse at t = 0 and t = 4 s? (Consider one-dimensional motion only.)


Suppose you are running fast in a field and suddenly find a snake in front of you. You stop quickly. Which force is responsible for your deceleration?


A free 238U nucleus kept in a train emits an alpha particle. When the train is stationary, a nucleus decays and a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x at time t after the decay. If the decay takes place while the train is moving at a uniform velocity v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger, is


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?


In the following figure, m1 = 5 kg, m2 = 2 kg and F = 1 N. Find the acceleration of either block. Describe the motion of m1 if the string breaks but F continues to act.


A block A can slide on a frictionless incline of angle θ and length l, kept inside an elevator going up with uniform velocity v in the following figure. Find the time taken by the block to slide down the length of the incline if it is released from the top of the incline.


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.


A body of mass m moving with a velocity v is acted upon by a force. Write an expression for change in momentum in each of the following cases: (i) When v << c, (ii) When v → c and (iii) When v << c but m does not remain constant. Here, c is the speed of light.


Two balls A and B of masses m and 2 m are in motion with velocities 2v and v, respectively. Compare:

(i) Their inertia.

(ii) Their momentum.

(iii)  The force needed to stop them in the same time.


State Newton's second law of motion. Under what condition does it take the form F = ma?


Use Newton's second law of motion to explain the following instance : 

A cricketer pulls his hands back while catching a fast moving cricket ball .


A force of 10 N acts on a body of mass 2 kg for 3 s, initially at rest. 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 stone is dropped freely from the top of a tower and it reaches the ground in 4 s. Taking g = 10m s-2, calculate the height of the tower.


A motorcycle of mass 100 kg is running at 10 ms−1. If its engine develops an extra linear momentum of 2000 Ns, calculate the new velocity of a motorcycle.


A ball is thrown upward and reaches a maximum height of 19.6 m. Find its initial speed?


A stone is dropped from a tower 98 m high. With what speed should a second stone be thrown 1 s later so that both hit the ground at the same time?


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)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×