Physics help?

Question

A woman at an airport is towing her 20.0 kg suitcase along a horizontal surface for 10 m from rest, she is pulling a constant 40 N force along the strap which is at an angle 30 degrees with respect to the horizontal. The coefficient of kinetic friction between the suitcase and the floor is .150.

a.) what normal force does the ground exert on the suitcase?
b) how much work is done by the pulling force?
c) How much work is done by the friction force?:
d) what is the speed of the suitcase after 10 m? You can use kinematics or work KE theorem.

I tried working out the problem but need to know if I doing it right.

For a, I used the formula Fy + n - w = 0 to find normal force.
40sin(30) + n - (9.8 x 20) = 0
20 + n - 196 = 0
n = 176 N

For b, I used the formula w = F sin (theta) (delta x) and got
w = (35) sin (30) (10) = 175 J

For c, I used friction = n (uk) where n = mg - F(app)sin(theta)
(9.8)(20) - 40 sin (30)
196 -20 = 176 (.150) = 26.4 J

I need help with starting d, also. Thnks

Answer 1

Hey, I havnt taken physics in a while, but I'm working on my masters in biochem, and i had minor in math in my undergrad, so I should be able to help a little.

a) The normal force is just the weight of the object - the weight she is pulling up with. So yup, you did it right!

b) Ok, I'm not positive, but I think its talking about Horizontal force, cause its not moving up and down, so you need to use cos instead of sin. This gives you an answer closer to 346 J. Not sure tho.

c) Im pretty sure you did this one completely wrong, but again, not sure. You're right up to where you get 26.4 as an answer, but that would be in Newtons, not Joules. You just discovered the Force of friction acting against motion. So thats a constant force acting over the 10m the suitcase was displaced. So, as you correctly stated in b) w = F sin (theta) (delta x), where theta would be 0 because its being dragged along the ground, deltax = 10 and F is the 26.4N. So the work done by friction is 10 x 26.4 = 264J

d) You can use either kinematics, or kinetic energy formulations, so KE = 1/2mv^2, or

V(final) = V(initial) + 2a(delta(x))
AND F=ma, so a=F/m, which turns the above into:
V^2(final) = V^2(initial) + [2F(delta(x))] / m

either way,
(346 - 264) = 1/2(20kg)(v^2) (gotta subtract energy lost to friction from the total energy of the object)
82 = 10v^2
v^2=8.2
v = 2.86m/s

OR

v^2 = 0 + [(2)(34.6 - 26.4)(10)] / 20
v^2 = 8.2
v = 2.86m/s

Use whichever one (or both) you feel like:) Hope this helps!

Answer 2

I am not going to check your previous answers, but d is actually pretty easy.

The Kinetic Energy is your answer to b) minus your answer to c).

KE=1/2 mv^2.

Solve for V.

Answer 3

I think(key word)
Take the work by pulling force - friction force work
---------------------- 175J-26.4J=148.6J
then work that backwards in the w=Fsin(theta)(delta x) equation to find a new force value. use this new force/mass to find acceleration. and lastly v^2=2ax

not sure, but it seems to make sense...

Answer 4

I tried. Email my buddy ..me first. He has a b.s. in this and helped me through it. He loves it.