A 0.296-kg block resting on a frictionless, horizontal surface is attached to a
ID: 1462089 • Letter: A
Question
A 0.296-kg block resting on a frictionless, horizontal surface is attached to a spring having force constant 83.8 N/m as in the figure below. A horizontal force F causes the spring to stretch at a distance of 5.33 cm from its equilibrium position.
(a) Find the value of F. (Enter the magnitude of the force only.) N
(b) What is the total energy stored in the system when the spring is stretched? J
(c) Find the magnitude of the acceleration of the block immediately after the applied force is removed. m/s2
(d) Find the speed of the block when it first reaches the equilibrium position. m/s
(e) If the surface is not frictionless but the block still reaches the equilibrium position, how would your answer to part (d) change? The block would arrive at a greater speed. The block would arrive at a lower speed. The block would arrive at the same speed.
(f) What other information would you need to know to find the actual answer to part (d) in this case?
Explanation / Answer
m =0.296 kg , k =83.8 N/m , x =5.33 cm
(a) F =kx =(83.8*0.0533)
F = 4.47 N
(b) E =(1/2)kx^2 =(1/2)(83.8)(0.0533*0.0533)
E = 0.119 J
(c) F =kx = ma
a = F/m = 4.47/0.296
a = 15.1 m/s2
(d) From conservation of energu
K =U
(1/2)mv^2 =(1/2)kx^2
v = (k/m)1/2x
v = [83.8/0.296]1/2(0.0533)
v = 0.897 m/s
(e) If the surface have friciton, then the velocity at equllibrium position is decreased.
The block arrive at lower sppeed.
(f) coefficienct of kinetic fricition force
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