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Question

A 36 kg crate, originally at rest, is dragged across a floor by pulling on a rope inclined 15° above the horizontal.The coefficients of static and kinetic friction are respectively µs=0.62 and µk = 0.53, and the tension in the rope is F=222 N.
(a) What is the magnitude of the normal force?
N
(b) What is the minimum horizontal force needed to start the crate?
N
(c)What is the magnitude of the friction force acting on the crate?
N
(d) What is the magnitude of the acceleration of the crate?
m/s2
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A 40.0 kg wagon is towed up a hill inclined at 18.8° with respect to the horizontal. The tow rope is parallel to the incline and has a tension of 140 N in it. Assume that the wagon starts from rest at the bottom of the hill, and neglect friction. How fast is the wagon going after moving 75 m up the hill?
m/s
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Answer 1

Consider the forces on the crate:

gravity = m*g
Friction = N*u
could be static or kinetic

the rope 222 N which has two components:

Horizontal:
Cos(15)*222
Vertical:
Sin(15)*222

a) The normal force is the resultant vertical forces on the block:
N=30*9.8-Sin(15)*222
=236.54N

b) Horizontal force to overcome static friction is
=236.54*.62
=146.65N
c) the horizontal component of the rope tension is
214.44N
so the crate is in motion.
The frictional force is the normal times the coefficient of kinetic friction
=.53*236.54N
=125.37N
The difference of the horizontal component of the rope is accelerating the crate:
F=m*a
a=(214.44-125.37)/30
a=2.97m/s^2
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for the second problem I'm going to use conservation of energy.

The work done by the tow rope = F*d
there is also potential energy gain
m*g*sin(18.8)*d
The difference will be in kinetic energy:
1/2*m*v^2

F*d-m*g*sin(18.8)*d=
.5*m*v^2
v=sqrt((F*d-m*g*sin(18.8)*d)*2/m)

plug in all of the numbers to get

v=sqrt((140*75-40*9.8*sin(18.8)*75)*2/40)
=7.16 m/s

j