A person pushes a large 42.9 kg box at a constant velocity of 9 m/s across a horizontal floor for 3.8 s. Find the work done ?
(given that the coefficient of kinetic friction Hk = 0.7)

Work =

10 kJ

Explanation:

W = Fd

W = (μN)(vt)

W = μ(mg)vt

W = 0.7(42.9)(9.81)(9)(3.8)

W = 10,075.12506 J

W ≈ 10 kJ

i wanna say it’s between b and d, leaning more towards b.

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