Inl (8c8p29) A 2.00 kg block is placed against a spring on a frictionless 24° in
ID: 1782721 • Letter: I
Question
Inl (8c8p29) A 2.00 kg block is placed against a spring on a frictionless 24° incline. The spring, whose spring constant is 19.8N/cm, is compressed 24.6 cm and then released. What is the elastic potential energy of the compressed spring? Submit Answer Tries 0/12 What is the change in the gravitational potential energy of the block-Earth system as the block moves from the release point to its highest point on the incline? Submit Answer Tries 0/12 How far along the incline is the highest point from the release point? Submit Answer Tries 0/12Explanation / Answer
given
m = 2 kg
k = 19.8 N/cm = 1980 N/m
x = 24.6 cm = 0.246 m
A) elastic poetntial energy stored in the spring,
U = (1/2)*k*x^2
= (1/2)*1980*0.246^2
= 60 J
B) change in gravitational potential energy = -loss of elastic pitential energy
= 60 J
C) let d is the distance travelled by the block along the incline bfore coming to rest.
Apply, (1/2)*k*x^2 = m*g*d*sin(theta)
60 = m*g*d*sin(24)
d = 60/(m*g*sin(24))
= 60/(2*9.8*sin(24))
= 7.53 m