A 0.200-kg block resting on a frictionless, horizontal surface is attached to a
ID: 1462572 • Letter: A
Question
A 0.200-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 5.32 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/s^2 (d) Find the speed of the block when it first reaches the equilibrium position. E 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 = 0.2 kg
spring constant k= 83.8 N/m
strech x=0.0532m
a)F= kx = 83.8 * 0.0532 = 4.458N
b) total enrgy E = 1/2 * k x^2
= 0.5 * 83.8 * 0.0532 ^2
= 0.0118 J
c) acceleration a=k*x/m = 83.8 *0.0532/ 0.2
a = 22.29 m/s^2
d) speed of the block at equillibrium is from
1/2 m v^2 = 0.0118
v= sqrt(2*0.0118)/0.2
v= 1.08 m/s
e) the block would arrive at lower speed
f)the additional information we need is frictional force
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