- EXAMPLE 5.9 A Horizontal Spring GOAL Use conservation of energy to calculate t

Question

- EXAMPLE 5.9 A Horizontal Spring GOAL Use conservation of energy to calculate the speed of a block on a horizontal spring with and without friction. PROBLEM A block with mass of 5.00 kg is attached to a horizontal spring with spring constant k = 4.00 × 102 N/m, as in the figure. The surface the block rests upon is frictionless. If the block is pulled out to xi = 0.0500 m and released, (a) find the speed of the block when it first reaches the equilibrium point, (b) find the speed when x = 0.0250 m, and (c) repeat part (a) if friction acts on the block, with coefficient = 0.150. 0 A mass attached to a spring. STRATEGY In parts (a) and (b) there are no nonconservative forces, so conservation of energy, can be applied. In part (c) the definition of work and the work-energy theorem are needed to deal with the loss of mechanical energy due to friction. SOLUTION (A) Find the speed of the block at the equilibrium point. (KE + PEg + PEs): = (KE + PEg + PEs)f Start with the conservation of energy equation Substitute expressions for the block's kinetic energy and the potential energy, and set the gravity terms to ero. Substitute vi = 0, xf = 0, and multiply by 2/m. Solve for vy and substitute the given Vf = V m ' ,-1 4,00 × 10-N/m (0.0500 m) values 5.00 kg = 0.447 m/s (B) Find the speed of the block at the halfway point. set v, o in Equation (1) and multiply by 2/m

Explanation / Answer

(a)

From conservation of energy,

(1/2)kx^2 = (1/2)mv^2

4.14*10^2*(0.046)^2 = 5.55*v^2

v = 0.297 m/s

(b)

(1/2) mv'2 = (1/2 )kx12 - 1/2 kx22

5.55*v'^2 = 4.14*10^2*(0.046^2 - 0.021^2)

v' = 0.353 m/s

(c)

From conservation of energy,

Total kinetic energy = total PE

(1/2)mv^2 = (1/2)kx^2 - umgx

where, umgx = work done by friction force

(1/2)*5.55*v^2 = (1/2)*414*0.046^2 - 0.16*5.55*9.8*0.046

v = 0.116 m/s

2. (a)

Let, distnce travelled b block = x

(1/2)kx^2 = (1/2)mv^2

(1/2)*414*x^2 = (1/2)*5.55*0.55^2

x = 0.0636 m

(b)

As the block is coming to rest, work done by friction force will add with PE of spring.

(1/2)mv^2 = (1/2)kx^2 + umgx

(1/2)*5.55*0.55^2 = (1/2)*414*x^2 + 0.16*5.55*9.8*x

207x^2 + 8.7024x - 0.839 = 0

by solving the quadratic equation,

x = 0.04604 m

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