(a) The work done on the block by the gravitational force is equal to the change in the gravitational potential energy of the block. The gravitational potential energy of the block is equal to mgh, where m is the mass of the block (250 g), g is the acceleration due to gravity (9.8 m/s2), and h is the height from which the block is dropped. Since the block is dropped from rest, the change in gravitational potential energy is equal to mgh. Therefore, the work done on the block by the gravitational force is equal to 250 g x 9.8 m/s2 x 12 cm = 294 J.
(b) The work done on the block by the spring force is equal to the change in the elastic potential energy of the spring. The elastic potential energy of the spring is equal to 1/2 kx2, where k is the spring constant (2.5 N/cm) and x is the compression of the spring (12 cm). Therefore, the work done on the block by the spring force is equal to 1/2 x 2.5 N/cm x (12 cm)2 = 180 J.
(c) The speed of the block just before it hits the spring can be calculated using the equation v2 = 2gh, where v is the speed of the block, g is the acceleration due to gravity (9.8 m/s2), and h is the height from which the block is dropped (12 cm). Therefore, the speed of the block just before it hits the spring is equal to √2 x 9.8 m/s2 x 12 cm = 17.3 m/s.
Problem 4:
A 250 g block is dropped onto a relaxed vertical spring that has a spring constant of k= 2.5 N/cm. The block becomes attached to the spring and compresses the spring 12 cm before momentarily stopping. While the spring is being compressed,
(a) what work is done on the block by the gravitational force on it?
(b) what work is done on the block by the spring force?
(c) what is the speed of the block just before it hits the spring? (neglect friction).
1 answer
2 answers