s = 1/2 at^2, so
1/2 a * 2.29^2 = 2
a = -0.76 m/s^2
F = ma
work = F*d
v^2 = 2as, and initial KE = 1/2 mv^2
Suppose you push a box, which is then slowed to rest by friction.
Given displacement = 2m, time = 2.29s, and mass of box = 300 g,
How do you find Work done by friction to stop the box & initial kinetic energy
1 answer