A steel cable lying flat on the floor drags a 20 kg block across a horizontal, frictionless floor. A 100 N force applied to the cable causes the block to reach a speed of 4.2 m/s in a distance of 2.0 m. What is the mass of the cable?

m_cable = 2,676 kg

Explanation:

For this exercise we must look for the acceleration with the kinematic ce relations

          v² = v₀² + 2 a x

since the block starts from rest, its initial velocity is vo = 0

          a = v² / 2x

          a = 4.2² /(2 2.0)

          a = 4.41 m / s²

now we can use Newton's second law

Note that the mass that the extreme force has to drag is the mass of the block plus the mass of the cable.

          F = (m + m_cable) a

          m_cable  = F / a -m

          m_cable = 100 / 4.41 - 20

          m_cable = 2,676 kg

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