(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
(b) The net work done on the crate while it is on the rough surface is 23.7 J.
(c) The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
<h3>Magnitude of net force on the crate</h3>
F(net) = F - μFf
F(net) = 280 - 0.351(92 x 9.8)
F(net) = -36.46 N
<h3>Net work done on the crate</h3>
W = F(net) x L
W = -36.46 x 0.65
W = - 23.7 J
<h3>Acceleration of the crate</h3>
a = F(net)/m
a = -36.46/92
a = - 0.396 m/s²
<h3>Speed of the crate</h3>
v² = u² + 2as
v² = 0.845² + 2(-0.396)(0.65)
v² = 0.199
v = √0.199
v = 0.45 m/s
Thus, the magnitude and direction of the net force on the crate while it is on the rough surface is 36.46 N, opposite as the motion of the crate.
The net work done on the crate while it is on the rough surface is 23.7 J.
The speed of the crate when it reaches the end of the rough surface is 0.45 m/s.
Learn more about work done here: brainly.com/question/8119756
#SPJ1
