English
Karnataka Board PUCPUC Science Class 11

Two masses 8 kg and 12 kg are connected at the two ends of a light, inextensible string that goes over a frictionless pulley. Find the acceleration of the masses,

Advertisements
Advertisements

Question

Two masses 8 kg and 12 kg are connected at the two ends of a light, inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Numerical
Advertisements

Solution

The given system of two masses and a pulley can be represented as shown in the following figure:

Smaller mass, m1 = 8 kg

Larger mass, m2 = 12 kg

Tension in the string = T

Mass m2, owing to its weight, moves downward with acceleration a,and mass m1moves upward.

Applying Newton’s second law of motion to the system of each mass:

For mass m1:

The equation of motion can be written as:

T – m1g = ma … (i)

For mass m2:

The equation of motion can be written as:

m2g – T = m2a … (ii)

Adding equations (i) and (ii), we get:

(m_2-m_1)g = (m_1+ m_2)a

`:.a = ((m_2-m_1)/(m_1+m_2))g ...(iii)`

`=((12-8)/(12+8)) xx 10= 4/20 xx 10 = 2 "m/s"^2`

Therefore, the acceleration of the masses is 2 m/s2.

Substituting the value of a in equation (ii), we get:

`m_2g - T = m_2 ( (m_2-m_1)/(m_1+m_2))g`

`=((2m_1m_2)/(m_1+m_2))g`

`=(2xx12xx8)/(12+8)xx10`

`=(2xx12xx8)/20 xx 10  = 96 N`

Therefore, the tension in the string is 96 N.

shaalaa.com
  Is there an error in this question or solution?
Chapter 4: Laws of Motion - EXERCISES [Page 70]

APPEARS IN

NCERT Physics [English] Class 11
Chapter 4 Laws of Motion
EXERCISES | Q 4.16 | Page 70

RELATED QUESTIONS

A body of mass 0.40 kg moving initially with a constant speed of 10 m s–1 to the north is subject to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0, the position of the body at that time to be x = 0, and predict its position at t = –5 s, 25 s, 100 s.


A helicopter of mass 1000 kg rises with a vertical acceleration of 15 m s–2. The crew and the passengers weigh 300 kg. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.


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?


Two objects A and B are thrown upward simultaneously with the same speed. The mass of A is greater than that of B. Suppose the air exerts a constant and equal force of resistance on the two bodies.


In a TV picture tube, electrons are ejected from the cathode with negligible speed and they attain a velocity of 5 × 106 m/s in travelling one centimetre. Assuming straight-line motion, find the constant force exerted on the electrons. The mass of an electron is 9.1 × 10−31 kg.


Consider the situation shown in the following figure. Both the pulleys and the string are light and all the surfaces are frictionless.

  1. Find the acceleration of the mass M.
  2. Find the tension in the string.
  3. Calculate the force exerted by the clamp on the pulley A in the figure.


Find the acceleration of the block of mass M in the situation shown in the following figure. All the surfaces are frictionless and the pulleys and the string are light.


Find the acceleration of the 500 g block in the following figure.


Two bodies A and B of same mass are moving with velocities v and 2v, respectively. Compare their (i) inertia and (ii) momentum.


The unit of linear momentum is :


A pebble is dropped freely in a well from its top. It takes 20 s for the pebble to reach the water surface in the well. Taking g = 10 m s-2 and speed of sound = 330 m s-1. Find : The depth of water surface


Name the physical entity used for quantifying the motion of a body.


Which of the following has the largest inertia?


Use Newton's second law to explain the following:
We always prefer to land on sand instead of hard floor while taking a high jump.


The INCORRECT statement about Newton's second law of motion is


A body of mass 2 kg travels according to the law x(t) = pt + qt2 + rt3 where p = 3 ms−1, q = 4 ms−2 and r = 5 ms−3. The force acting on the body at t = 2 seconds is ______.


Newton’s Second Law corrected the old belief that force is needed to maintain motion. This idea came from the philosopher:


Why is catching a slow-moving ball easier than catching a fast-moving ball?


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×