Question
after a spring is compressed 0.2 meters from its uncompressed length it exerts a 20 newton force on a 0.4 kg block resting on a horizontal frictionless surface. after it is released the block will reach a maximum velocity of
a. 2.1 m/s
b. 3.2 m/s
c. 6.3 m/s
d. 10.0 m/s
e. 100 m/s
a. 2.1 m/s
b. 3.2 m/s
c. 6.3 m/s
d. 10.0 m/s
e. 100 m/s
Answers
R_scott
spring constant ... k = 20 N / 0.2 m
stored energy = 1/2 k x^2 = 1/2 * 100 * 0.2^2 Joules
the K.E. of the block equals the stored energy in the spring
1/2 m v^2 = 2 J
v^2 = 4 J / 0.4 kg
stored energy = 1/2 k x^2 = 1/2 * 100 * 0.2^2 Joules
the K.E. of the block equals the stored energy in the spring
1/2 m v^2 = 2 J
v^2 = 4 J / 0.4 kg