Home
Class 11
PHYSICS
An ideal massless spring S can compresse...

An ideal massless spring `S` can compressed `1.0` m in equilibrium by a force of `1000 N`. This same spring is placed at the bottom of a friction less inclined plane which makes an angle `theta =30^@` with the horizontal. A (10 kg) mass (m) is released from rest at the top of the incline and and is brought to rest momentarily after compressing the spring by `2 m`. If `g =10 ms^(-1)`, what is the speed of just before it touches the spring? .

A

`sqrt20 ms^(-1)`

B

`sqrt30 ms^(-1)`

C

`sqrt10 ms^(-1)`

D

`sqrt40 ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
`F=kx`
`:. k=(F)/(x) =(100)/(1)`
`=100 N//m`
`E_(i)=E_(f)`
`:. 1/2 xx 10 xx v^(2) =1/2 xx 100 xx (2)^(2)-(10)(10)(2 sin 30^@)`
Solving we get,
`v=sqrt20 m//s`
Promotional Banner

Similar Questions

Explore conceptually related problems

An ideal massless spring S can compressed 1.0 m in equilibrium by a force of 1000 N . This same spring is placed at the bottom of a friction less inclined plane which makes an angle theta =30^@ with the horizontal. A 10 kg mass m is released from the rest at top of the inclined plane and is brought to rest momentarily after compressing the spring by 2.0 m. the distance through which the mass moved before coming to rest is.

An ideal massles spring S can be compressed 1 m by a force of 100 N in equilibrium. The same spring is placed at the bottom of a frictionless plane inclined at 30^(@) to the horizontal. A 10 kg block M is released from rest at the top of the incline and is brought to rest momentarily after compressing the spring by 2 m. If g = 10 m//s^(2) , the speed of mass just before it touches the spring is sqrt(10x) m//s . Find value of x?

An ideal massless spring is compressed by 1 m by a force of 100 N. The same spring is placed at the bottom of a frictionless inclined plane (theta = 30^(@)) . A 10-kg mass is released from the top of the incline and is brought to rest momentarily after compressing the spring by 2m. Through what distance does the mass slide before coming to rest ? [Hint : Apply work-energy theorem]

When a spring is compressed by 0.05 m a restoring force of 10 N is developed in it. Then the force constant of the spring is

A block of 4kg mass starts at rest and slides a distance d down a friction less incline ( angle 30^@ ) where it runs into a spring of negligible mass. The block slides an additional 25m before it is brought to rest momentarily by compressing the spring. The force constant of the spring is 400Nm^-1 . The value of d is (take g=10ms^-2 )

A block of mass sqrt2 kg is released from the top of an inclined smooth surface as shown in fighre. If spring constant of spring is 100 N/m and block comes to rest after compressing the spring by 1 m, then the distance travelled by block befor it comes to rest is

In an ideal pulley particle system, mass m_2 is connected with a vertical spring of stiffness k . If mass m_2 is released from rest, when the spring is underformed, find the maximum compression of the spring.

An inclined plane is inclined at an angle theta with the horizontal. A body of mass m rests on it, if the coefficient of friction is mu , then the minimum force that has to be applied to the inclined plane to make the body just move up the inclined plane is

A block of mass m initially at rest is dropped from a height h on to a spring of force constant k . the maximum compression in the spring is x then

The block shown in figure is released from rest. Find out the speed of the block when the spring is compressed by 1 m