MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Work Energy Theorem Application

Two blocks of...

Question

Two blocks of masses $M_{1}$ and $M_{2}$ are connected by spring of constant $K$ . The spring is initially compressed and the system is released from rest at $t = 0$ second. The work done by spring on the blocks $M_{1}$ and $M_{2}$ be $W_{1}$ and $W_{2}$ respectively by time t. The speeds of both the blocks at time t are non zero. Then the value of $\frac{W_{1}}{W_{2}}$ equals to

A

\frac{M_{1}}{M_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

B

\frac{M_{2}}{M_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C

{(\frac{M_{1}}{M_{2}})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

D

{(\frac{M_{2}}{M_{1}})}^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{M_{2}}{M_{1}}$
$W_{1} = \frac{1}{2} M_{1} V_{1}^{2} W_{2} = \frac{1}{2} M_{2} V_{2}^{2} M_{1} V_{1} = M_{2} V_{2} \frac{W_{1}}{W_{2}} = \frac{M_{1} V_{1}}{M_{2} V_{2}} . \frac{V_{1}}{V_{2}} = \frac{M_{2}}{M_{1}}$

Suggest Corrections

thumbs-up

0

Similar questions

(m_{2} > m_{1})

are held in such a manner that spring is unstretched initially. Consider the situation at the time of release of the block

(t = 0)

and short time after releasing the block

(t > 0)

. Accelerations of the blocks of masses

m_{1}

m_{2}

a_{1}

a_{2}

Q. Two blocks of mass

m_{1}

m_{2}

(m_{1} < m_{2})

are connected with an ideal spring on a smooth horizontal surface as shown in figure. At

t = 0

m_{1}

m_{2}

is given a velocity

v

towards right. At this moment, spring is in its natural length. Then choose the correct alternative:

(Assume all surfaces are smooth)

m_{1}

m_{2}

are connected by a massless spring of constant

k

and unstretched length

L

. The masses are placed on a frictionless straight channel, which is considered as x -axis. They are initially at rest at

x = 0

x = L

respectively. At

t = 0

, a velocity of

v_{0}

is suddenly imparted to the first particle. At a later time

t_{0}

, the centre of mass of the two masses is at

x =

Q. Two blocks of masses

m_{1}

m_{2}

are connected by a massless spring and placed on smooth surface. The spring initially stretched and released. Then :

Q. The two blocks of masses

m_{1}

m_{2}

are kept on a smooth horizontal table as shown in the figure. Block of mass

m_{1}

m_{2}

is fastened to the spring. If now both the blocks are pushed to the left, so that the spring is compressed at a distance

d

. The amplitude of oscillation of block of mass

m_{1}

after the system released, is :

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Conservative Forces and Potential Energy

PHYSICS

Watch in App

explore_more_icon

Explore more

Work Energy Theorem Application

Standard XII Physics

Join BYJU'S Learning Program

CrossIcon