One end of a horizontal spring with the spring constant 1900 N/m is attached to the wall, the other end is attached to a block of mass 1.15 kg. Initially, the spring is compressed 4.5 cm. When released, the spring pushes the block away and is no longer in contact with the block. The block slides along a horizontal frictionless plane. a/ Compute the maximum speed of the block.
b/ The block goes off the edge of the plane and falls down from the plane to reach the floor with speed of
7 m/s. How high is the plane with respect to the floor?

(a) When the spring is compressed 4.5 cm = 0.045 m, it exerts a restoring force on the block of magnitude

F = (1900 N/m) (0.045 m) = 85.5 N

so that at the moment the block is released, this force accelerates the block with magnitude a such that

85.5 N = (1.15 kg) a   ==>   a = (85.5 N) / (1.15 kg) ≈ 74.3 m/s²

The block reaches its maximum speed at the spring's equilbrium point, and this speed v is such that

v ² = 2 (74.3 m/s²) (0.045 m)   ==>   v = √(2 (74.3 m/s²) (0.045 m)) ≈ 2.59 m/s

(b) There is no friction between the block and plane, so the block maintains this speed as it slides over the edge. At that point, it's essentially in free fall, so if y is the height of the plane, then

(7 m/s)² - (2.59 m/s)² = 2gy   ==>   y = ((7 m/s)² - (2.59 m/s)²) / (2g) ≈ 2.16 m

compass can be used day and night, cloudy or sunny

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s