# What is the wagon's displacement when t=7.0s?

Superwoman was seen packing up a suspicious bundle into a wagon, which has a total mass of 25.0 kg and is at rest. She pulls the wagon towards the east with an applied force of 60.0 N, which makes an angle of 35.0° above the horizontal. The coefficient of friction is 0.100 between the wagon wheel and the ground.
a) Draw and label a free body diagram.
b) Calculate the acceleration of the wagon.
c) If the acceleration remains constant, what will the final velocity of the wagon be when t = 7.0s?
d) If the work done by air resistance is 15J, determine the final velocity of the wagon when t=7.0s.
e) What is the wagon's displacement when t=7.0s?

So I drew a FBD. And found that a=1.12 m/s^2 (by finding Fnormal using Fnety = Fnormal + Fapplied - Fgravity = 0 and then using Fnormal in Fnetx = Fapplied*cos35 - Ffriction. Then Fnetx = ma)

c) V=Vo +at
t = 7.0s
Vo = 0
a = 1.12
Therefore V = 7.9 m/s

d) ?? Here's the answer the teacher gave me, but I don't understand it:
1/2mv^2 -15 = 1/2mv2^2
1/2(25)(7.9)^2 -15 = 1/2(25)(V2^2)
765.125 = 1/2(25)(V2^2)
V2 = 7.82 m/s [E]

I don't get why the air resistance is subtracted from Work done at t=7.0s. Please help explain part d to me...thank you!

By definition of work, $$W=\Delta E= \frac{1}{2}m\Delta\left( v^2 \right) = \frac{1}{2}mv_{f}^2 - \frac{1}{2}mv_{o}^2$$
Since you are giving W, then you solve for $$v_f$$.

Last edited:
nicksauce