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# A block lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 57 N/m. ?

A block 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 2.9 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.

(a) What is the position of the block?

(b) What is the work that has been done on the block by the applied force?

(c) What is the work that has been done on the block by the spring force?

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

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

(e) The value of that maximum kinetic energy.

### 1 Answer

- NCSLv 73 months agoFavorite Answer
(a) At the stopping point (which is BEYOND the equilibrium point due to the force), the work done by the force is equal to the work done by the spring:

Fx = ½kx²

x = 2F / k = 2*2.9N / 57N/m = 0.10 m

(b) W = ½kx² = ½ * 57N/m * (0.10m)² = 0.30 J

(c) Since the KE is zero, W' = -0.30 J

(d) KE(x) = Fx - ½kx²

At maximum, dKE/dx = 0 = F - kx

x = F / k = 2.9N / 57N/m = 0.051 m

(e) At that point, SPE = ½ * 57N/m * (0.051m)² = 0.074 J

and work done W = Fx = 2.9N * 0.051m = 0.15 J

and so KE = W - SPE = 0.15 - 0.074J = 0.074 J

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