A 0.294-kg box is placed in contact with a spring of stiffness 3.792×102 N/m. The spring is compressed 1.42×10-1 m from its unstrained length. The spring is then released, the block slides across a frictionless tabletop, and it flies through the air. The tabletop is a height of 0.960 m above the floor.

Question

1. What is the potential energy stored in the spring when it is compressed?

2. What is the kinetic energy of the block just before it leaves the table but after it is no longer in contact with the spring?

3. What is the kinetic energy of the block just before it hits the floor?

4. How much time elapses between the time when the block leaves the table and the time just before the block hits the floor?

Answer 1

1.
U = (1/2) k x^2
=(1/2)(3.792*10^2) (1.42^2)(10^-2)
= 3.82 Joules

2. same 3.82 Joules

3. 3.82 + m g h
= 3.82 + .294(9.81)(0.960)
= 6.59 Joules

4. How long does it take to fall .96 meters?

.96 = (1/2)(9.81)(t^2)

Answer 2

Thank you so much! And What distance, d, from the edge of the table does the block hit the floor?

Answer 3

I keep getting .154

Answer 4

KE=1/2 m v^2
v= sqrt(2KE/m)=sqrt(2*3.82/.294)=5.10 m/s
distance=velocity*time=5.1(time)
time=sqrt(2*.96/9.81)=.442sec

h=5.1*.442 meters

Answer 5

Thank you!