CameraIcon

SearchIcon

MyQuestionIcon

3

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Second Law of Motion

Consider the ...

Question

Consider the system shown in figure. System is released from rest, find the tension (in Newton) in the cord connected between 1 kg and 2 kg blocks. $(g = 10 \frac{m}{s^{2}})$

Open in App

Solution

By Newton's second law, we have
$1 g - T_{1} = 1 a . . . . . . . . . . . (i)$
$(T_{1} + 2 g) - T_{2} = 2 a . . . . . . . . . . . (i i)$
and $T_{2} - 2 g = 2 a . . . . . . . . . . . (i i i)$
Solving equation (i), (ii) & (iii), we get
$a = 2 \frac{m}{s^{2}}$
$T_{1} = 8 N$

Suggest Corrections

thumbs-up

0

Similar questions

Q. Consider the system shown in figure. System is released from rest, find the tension (in Newton) in the cord connected between 1 kg and 2 kg blocks.

(g = 10 \frac{m}{s^{2}})

Q. Two unequal masses of

1

2

kg are attached at the two ends of a light inextensible string passing over a smooth pulley as shown in figure. If the system is released from rest, find the work done by tension on both

1

2

1

g = 10 m / s^{2}

Q. The system shown is figure is released from rest. String is massless and pulley is smooth. Find the work done by string on 1 kg block in the time interval of

2 s

g = 10 m / s^{2}

)

Q. The system shown in figure is released from rest. Calculate the tension in the string connecting

3

1

kg, assuming pulleys and strings are masseless

m_{1} = 10 k g

m_{2} = 5 k g

are connected by an ideal string as shown in the figure. The coefficient of friction between

m_{1}

and the surface is

μ = 0.2

. Assuming that the system is released from rest. The velocity of blocks when

m_{2}

has descended by

4 m

(g = 10 m / s^{2})

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Second Law of Motion

PHYSICS

Watch in App

explore_more_icon

Explore more

Second Law of Motion

Standard XIII Physics

Join BYJU'S Learning Program

CrossIcon