# 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
Homework Helper
Meaning that W is the sum of all the done work done, ie work done on the wagon (pulling), and work done against the wagon (air resistance). Since the latter is done against the wagon, it is considered to be negative work.

Are both velocities for t=7.0s?

dynamicsolo
Homework Helper
Are both velocities for t=7.0s?

I'll butt in, since no one came back to this yet. Yes, it sounds like the question is stating that the total work done by air resistance is for those seven seconds. So the revised velocity in part (d) is also at t = 7.0 seconds.