Don't mean to spam this board, but I really need help with this last problem and I can't find anyone that can explain it to me.

The problem is as follows: This weekend my brother was doing yard work and decided to give our younger siblings a ride in the wheelbarrow, which was initially at rest, by pushing with a constant horizontal force of 10 Newtons on a surface with a coefficient of friction of 0.06. The mass of the wheelbarrow and kids is 55 kg. If my brother pushes the wheelbarrow for 3 seconds (starting from rest), what distance does the wheelbarrow cover during this time?

I think she (my professor) said that kinematics were needed to solve.

So basically, I'm given

force = 10 N
co-efficient of friction= 0.06
mass= 55 kg
time = 3 seconds

I keep drawing free-body diagrams, but I don't know what to do with the numbers I get.

There is an answer key that says the correct answer is 5.5 meters, but I need to know the steps taken to arrive at that answer.

dont worry, your question isnt spam, a lot of threads here is homework help
im going to assume you made a typo in your question. 10 N is a very small force and would not be able to overcome friction.

either that, or we are talking about rotational dynamics(for the wheel) and you havent given us things like radius of the wheel, or moment of inertia etc.

but i will assume youre talking about 100 N rather than 10 N

mass = 55 kg (given in question)

weight (force) = mass*gravity (acceleration)
weight = 55*9.81 (gravity is assumed to be 9.81 m/s^2)
weight = 539.55 N

normal force (the one that the ground pushes against the wheelbarrow) = weight
= 539.55 N

now this is only equal when the surface is horizontal. if it were on a slope, this force would be weight*cos(x) where x is the angle of elevation. since the angle is 0 on a flat surface, and cos(0) is 1, its just equal to the weight.

frictional force = normal force * coefficient of friction
frictional force = 539.55 * 0.06
frictional force = 32.373 N

notice that if you really had 10 N as the force, it wont be enough to overcome friction

applied horizontal force = 100 N (given in question)

friction acts in direction opposite to motion. so:
net force = applied force - friction
net force = 100 - 32.373
net force = 67.627

force = mass *acceleration
67.627 = 55 * acceleration
acceleration = 67.627/55
acceleration = 1.230 m/s^2

using your kinematic equations for constant acceleration
s = ut + 0.5at^2
where s = distance, u = initial velocity, t = time, a = acceleration.

u = 0 (starting from rest)

s = 0*3 + 0.5*1.230*3^2
s = 5.535 m

which is what you get in the answer

if asked to draw a free body diagram. obviously you have a wheelbarrow.

-there will be a weight force, which always goes downwards.
-there will be a normal force that is perpendicular to the surface (its opposite direction of the weight on a horizontal surface). this is the the one that stops the object from going into the ground.
-there will be the force you applied, going horizontally.
-there will be a frictional force going opposite to the applied force.

so basically, a 4 forces pointing up, down, left and right

================================================== ==
on an end note, im guessing youre doing fairly basic physics. this isnt really supposed to be the answer in real life. youll have to take into account the rotational dynamics of the wheel.
so really, its probably not a wheelbarrow, but maybe a box or something

in a physics test like this, i would recommend knowing
s = ut + 0.5at^2
v = u + at
v^2 = u^2 + 2as

where v = velocity at that point in time, s = displacement, u = initial velocity, a = acceleration (must be constant to work, like gravity), t = time in seconds

also, remember,

weight is a force = mass * acceleration (gravity = 9.81)
normal force = weight *cos(x) where x is the angle of elevation of a slope. cos (0), a horizontal surface , = 1
frictional force = normal force * coefficient of friction

friction always acts opposite of motion, so simply subtract it by the motion force.
force = mass * acceleration

also
im not sure if you will have to know this, but it is splitting the force into its x and y components. this is important for things like projectile motion as well as pulling on a wheelbarrow with a force that is not horizontal (eg with a force at 45 degrees). if so , the vertical component (and the one that will affect the weight, either increasing it or decreasing it. this will then affect frictional force ) is F*sin (x) where x is the angle at which you pull it and F being the magnitude of the force. and the horizontal component (affects the motion) = F*cos (x)


good luck on your test.

thank you so much, you have in about ten minutes clarified a concept that days of physics classes failed to deliver to me.

and on totse no less.

behold the power of the internet, and thank you for the good luck.

p.s.- you were correct, it was a typo and the intended number was 100 N.

dont worry, your question isnt spam, a lot of threads here is homework help
im going to assume you made a typo in your question. 10 N is a very small force and would not be able to overcome friction.

either that, or we are talking about rotational dynamics(for the wheel) and you havent given us things like radius of the wheel, or moment of inertia etc.

but i will assume youre talking about 100 N rather than 10 N

mass = 55 kg (given in question)

weight (force) = mass*gravity (acceleration)
weight = 55*9.81 (gravity is assumed to be 9.81 m/s^2)
weight = 539.55 N

normal force (the one that the ground pushes against the wheelbarrow) = weight
= 539.55 N

now this is only equal when the surface is horizontal. if it were on a slope, this force would be weight*cos(x) where x is the angle of elevation. since the angle is 0 on a flat surface, and cos(0) is 1, its just equal to the weight.

frictional force = normal force * coefficient of friction
frictional force = 539.55 * 0.06
frictional force = 32.373 N

notice that if you really had 10 N as the force, it wont be enough to overcome friction

applied horizontal force = 100 N (given in question)

friction acts in direction opposite to motion. so:
net force = applied force - friction
net force = 100 - 32.373
net force = 67.627

force = mass *acceleration
67.627 = 55 * acceleration
acceleration = 67.627/55
acceleration = 1.230 m/s^2

using your kinematic equations for constant acceleration
s = ut + 0.5at^2
where s = distance, u = initial velocity, t = time, a = acceleration.

u = 0 (starting from rest)

s = 0*3 + 0.5*1.230*3^2
s = 5.535 m

which is what you get in the answer

if asked to draw a free body diagram. obviously you have a wheelbarrow.

-there will be a weight force, which always goes downwards.
-there will be a normal force that is perpendicular to the surface (its opposite direction of the weight on a horizontal surface). this is the the one that stops the object from going into the ground.
-there will be the force you applied, going horizontally.
-there will be a frictional force going opposite to the applied force.

so basically, a 4 forces pointing up, down, left and right

================================================== ==
on an end note, im guessing youre doing fairly basic physics. this isnt really supposed to be the answer in real life. youll have to take into account the rotational dynamics of the wheel.
so really, its probably not a wheelbarrow, but maybe a box or something

in a physics test like this, i would recommend knowing
s = ut + 0.5at^2
v = u + at
v^2 = u^2 + 2as

where v = velocity at that point in time, s = displacement, u = initial velocity, a = acceleration (must be constant to work, like gravity), t = time in seconds

also, remember,

weight is a force = mass * acceleration (gravity = 9.81)
normal force = weight *cos(x) where x is the angle of elevation of a slope. cos (0), a horizontal surface , = 1
frictional force = normal force * coefficient of friction

friction always acts opposite of motion, so simply subtract it by the motion force.
force = mass * acceleration

also
im not sure if you will have to know this, but it is splitting the force into its x and y components. this is important for things like projectile motion as well as pulling on a wheelbarrow with a force that is not horizontal (eg with a force at 45 degrees). if so , the vertical component (and the one that will affect the weight, either increasing it or decreasing it. this will then affect frictional force ) is F*sin (x) where x is the angle at which you pull it and F being the magnitude of the force. and the horizontal component (affects the motion) = F*cos (x)


good luck on your test.
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