Course Contents » ... » HW /> Mass on Spring Notes E Bookmark & Evaluate Communi
ID: 2031486 • Letter: C
Question
Course Contents » ... » HW /> Mass on Spring Notes E Bookmark & Evaluate Communicate e Print Into A 1872.0 g mass is on a horizontal surface with pk = 0.30, and is in contact with a massless spring with a force constant of 682.0 N/m which is compressed. When the spring is released, it does 13.64 J of work on the mass while returning to its equilibrium position. Calculate the distance the spring was compressed. Submit Answer Tries 0/12 What is the velocity of the mass as it loses contact with the spring? Submit Answer Tries 0/12Explanation / Answer
< Calculate the distance the spring was compressed. >>
Spring energy = 13.64 = (1/2)Kx^2
where
K = spring constant = 682 N/m (given)
x = distance at which spring was compressed
Therefore,
13.46 = (1/2)(682)(x^2)
Solving for "x"
x = 0.198 m
<< what is the velocity of the mass as it breaks contact with the spring? >>
Energy from spring = Energy absorbed by block + Energy to overcome friction
Es = Eb + Ef
13.64= (1/2)mV^2 + µmgx
where
m = mass of the block = 1.872 kg. (given)
V = velocity at which block leaves the spring
µ = coefficient of friction = 0.30 (given)
g = acceleration due to gravity = 9.8 m/sec^2 (constant)
x = 0.198 m (as calculated above)
Substituting appropriate values,
13.64 = (1/2)(1.872)V^2 + 1.872(9.8)(0.198)
13.46 = 0.963V^2 + 3.632
9.828 = 0.963V^2
Solving for "V"
V = 3.194 m/sec.