4.82 kJ work is done by a worker pulling a crate across the deck of a ship at a constant speed of 7.82 m/s by a 453 N force. The rope is being pulled at a 37.0 degree angle to the deck.
a)How far was it pulled?
b)What is the friction force?
c)What is the work done against friction?
d)What is the coefficient of kinetic friction.
a)W=Fdcos0
4830J=454N(x)cos37
d=13.3m
b)Vconstant
Fx=0
Fappx=Ffk=454Ncos37
Ffk=363N
c)Wffk=Ffk(d)
=363N(13.3m)
=4828J
d)Ffk=uk(Fn)
uk=Ffk/mg
uk=363N/264.6
uk=1.37
That's the best I can come up with but the coefficient of kinetic friction is awfully high and I'm also not sure that I calculated properly with regards to the normal force. Does that decrease due to the presence of the y component of the F applied? Thank you very much in advance!
a. W = F*d = 453*Cos37 * d = 4820 J.
d = 13.3 m.
b. Fx-Fk = M*a
453*Cos37-Fk = M*0 = 0
Fk = 453*Cos37 = 361.8 N. = Force of kinetic friction.
c. W = Fk*d = 361.8 * 13.3 = 4812 J.
d. W = M*V^2/2 = 4820 J.
M*7.82^2/2 = 4820
M*30.58 = 4820
M = 157.6 kg = Mass of the crate.
M*g = 157.6 * 9.8 = 1545 N. = Wt. of the
crate. = Normal force(Fn).
u = Fk/Fn = 361.8/1545 = 0.234
Thank you Henry! You did a really nice job and it helped me a lot!! (:
Your calculations are mostly correct. Let's go through each part to understand the answers:
a) To calculate the distance pulled (d), you used the formula W = Fdcosθ, where W is the work done, F is the applied force, d is the distance, and θ is the angle between the force and the displacement. In this case, W is given as 4.82 kJ, F is 453 N, and θ is 37 degrees.
Plugging in the values, we have:
4830 J = 453 N * d * cos(37°)
Now, solve for d:
d = 4830 J / (453 N * cos(37°))
d ≈ 13.3 m
So, the crate was pulled a distance of approximately 13.3 meters.
b) Since the crate is moving at a constant speed, the net force along the horizontal direction (x-axis) must be zero. Therefore, the friction force (Ffk) must be equal in magnitude and opposite in direction to the applied force.
Ffk = Fappx = 453 N * cos(37°)
Ffk ≈ 363 N
So, the friction force is approximately 363 N.
c) The work done against friction (Wffk) can be found using the formula W = Fd, where F is the force and d is the distance. In this case, F is the friction force (363 N) and d is the distance pulled (13.3 m).
Wffk = Ffk * d
Wffk = 363 N * 13.3 m
Wffk ≈ 4828 J
Therefore, the work done against friction is approximately 4828 J.
d) To find the coefficient of kinetic friction (µk), you can use the equation µk = Ffk / (mg), where Ffk is the friction force and mg is the normal force. The normal force (Fn) is the force perpendicular to the surface.
In this case, the normal force is equal to the weight of the crate (mg). The weight (mg) is given by: mg = 453 N * sin(37°).
Now, calculate the coefficient of kinetic friction (µk):
µk = Ffk / (mg)
µk = 363 N / (453 N * sin(37°))
µk ≈ 1.37
So, the coefficient of kinetic friction is approximately 1.37.
Regarding your concern about the normal force, it does not decrease due to the presence of the y component of the applied force. The normal force is only affected by the weight (mg) of the crate, which acts vertically downward. The y component of the applied force does not have any influence on the normal force.