The block in the figure below lies on a horizontal frictionless surface and is a

Question

The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 57 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.6 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops. Assume that the stopping point is reached.

During the block's displacement, find the following values.

(d) The block's position when its kinetic energy is maximum.
m

(e) The value of that maximum kinetic energy.
mJ

Explanation / Answer

Block will stop when it's velocity (K.E = 0) reaches zero not when F=Kx. So block will stop when work done by force F is equal to potential energy of spring.

(a) Work done by the force = change in potential energy of the spring

F*x = 0.5*K*x2

2.6*x = 0.5*57*x2

x = 0.0912 m

(b) Since F is constant-

Work = F*x = 2.6*0.09122 = 0.2372 N

c) Work done by spring = change in potential energy of spring

change in P.E = - 0.5*K*x^2 = -0.5*57*0.091222

Work by spring = - 0.2371 J

d) Maximum kinetic energy is reached when Net force on block becomes zero.

let x be the displacement of block then,

F = K*x

2.6 = 57*x

x= 0.0456m

at 0.0456, Kinetic energy will be maximum.

e) Work done by spring+work done by Force = Change in Kinetic energy

-0.5*K*x2 + F*x = Kinetic energy max

K.E max= 0.05929 J = 59.29 mJ

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