Simple work and force physics question (highschool level)

Question

Question 17:A 20.0 kg wagon is pulled with a force of 40.0 N by its handle

at an angle of 30 degrees to the horizontal against a force of

friction of 20.0 N.

a) Determine the net force in the direction of motion.

(14.6 N)

b) Calculate the work done by the person in pulling the cart

10.0 m.

(346J)

c) If the wagon starts from rest, what is its speed after 10.0 m?

Remember that the acceleration is due to the net force.)

(3.83 m/s)

The answers are already here, I just want to learn the pross of finding the answer, thank you!

Answer 1

a)

First, draw a free body diagram. Split the force acting on the wagon into horizontal and vertical components. Since we know that the wagon will travel only horizontally, we need only the horizontal component of force:

40.0N*cos(30°)= 34.64 N

We can subtract from this the frictional force, since they are in opposite directions, to find the net force:

34.6 N-20.0N

F=14.6N

b)

From our free body diagram, we see that the person exerts a force of 34.6N. The work done is force times the component of the displacement in the direction of the force. Since they are parallel, this is

Work=force * distance=34.6N*10m=346N*m

Work=346J

c)

Here, we can apply conservation of energy by multiplying the net force acting on the wagon that we found earlier (14.6N) by the displacement in that direction. This will be equal to the kinetic energy of the wagon.

14.6N*10m = 146J

The kinetic energy of the wagon is related to its speed:

146J =½ m v²

v=sqrt(2*146J/20kg)

v=3.82m/s

Answer 2

Free body diagram

(imagine we are pulling the wagon to the right)

F pull = diagonally up at a 30 degree angle

Weight = down

Normal force = up

friction = opposite of motion = left

We need to break the pull force into x and y components so we can find the net force in the direction of motion. The direction of motion will include the x component of the pull force and the friction.

F pull = 40N hypotenuse points up and right

angle = 30 degrees

F pull x component = cosine

cosine = adjacent / hyp

cos(30) = Fpullx / 40

Fpullx = 40cos(30) = 34.6 N

Fpull y component = sine

sine = opposite/hyp

sin(30) = Fpully / 40

F pull y = 40sin(30) = 20 N

Now we know the forces in the direction of motion.

F pull x = 34.6 N to the right

Friction (given) = 20N to the left

Net force in the direction of motion = right - left

34.6 - 20

F net = 14.6N in direction of motion (to the right)

Work done by pulling = pulling force * distance (in same direction)

F pull in direction of motion = F pull x = 34.6N

distance= 10 meters

Are they in the same direction? Yes! So we can use this equation

Work = (34.6)(10) = 346 J

Net work = change in KE

Net work = net force * distance

net force = 14.6N

distance = 10 meters

net work done = (14.6)(10) = 146 J

The wagon started from rest and gained 146 J. The wagon started with 0 J of kinetic energy and ended with 146 J of kinetic energy. We can find the final KE by using

KE = 1/2mv^2

146 = 1/2mv^2

mass = 20kg

v=?

146 = 1/2(20)v^2

146 = 10v^2

14.6 = v^2 square root both sides

v = 3.8 m/s

this is the speed after pulling the wagon 10 meters