A block of mass m, lying on a smooth horizontal surface is attached to a spring of negligible masss of sprint constrant k.
The other end of the spring is fixed as shown in the figure.
The block is initiallY at rest in its equilibrium position if now the block is pulled with a conract force F THE MAXIMUM SPEED OF THE BLOCK IS
Answers
Answered by
17
by work energy theorem
W= change in K.E
wrk done by spring = 1/2 m{ (V2)^2- (V1)^2
1/2 K (x1^2-x2^2) = 1/2 m{ (V2)^2- (V1)^2
since initially block is at rest therefore, V1=0, and X1=0
so, by solving we will get
V= F/√mk
[change in k.E =]
[1/2 m{ (V2)^2- (V1)^2]
W= change in K.E
wrk done by spring = 1/2 m{ (V2)^2- (V1)^2
1/2 K (x1^2-x2^2) = 1/2 m{ (V2)^2- (V1)^2
since initially block is at rest therefore, V1=0, and X1=0
so, by solving we will get
V= F/√mk
[change in k.E =]
[1/2 m{ (V2)^2- (V1)^2]
Answered by
8
Answer:
Explanation:
According to conservation of energy we know that work done by a body is its final kinetic energy - initial kinetic energy.
W= kf - ki
W= 1/2* m * v^2 - 1/2* m * u^2
Where v is the final velocity and u is the initial velocity.
If the block is pulled by a distance x then force(F) = kx.
So, x=F/k .....(1)
Work done by F + Work done by the spring = Change in total kinetic energy.
2F.x - kx^2 = m*v^2 - 0.......(2)x1/2
F.x - 1/2 * kx^2 = 1/2 * m*v^2
Thus, from equation 1 and 2 we get:-
Putting the value of x in 2 we get.
F.F/k - 1/2 * k * (F/k)^2 = 1/2m*v^2
On solving we get.
v= F/✓(2mk).
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