A cart is loaded w/ bricks has a total mass of 28.1kg and is pulled @ constant speed by a rope. The rope is inclined @ 23 degrees above the horizontal & the cart moves 19.8m on a horizontal surface. The coefficient of kinetic friction between the ground and cart is 0.735. The acceleration of gravity is 9.8 m/s2.
2006-07-13 12:00:34 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The first assumption that is necessary is that the cart is not on wheels. Because, if it is on wheels, then the data provided in the problem is insufficient. This is because, if the wheels were rolling perfectly (no slip), then there would be no loss due to friction, hence all the work would be done in initially accelerating the cart to the desired speed. Of course, if there was no time limit, the best thing to do is to move infinitely slowly, then no work is required!
I find it that it is often more revealing to solve a problem using symbols before substituting numbers, so here it is:
F: force that the rope is pulled with = we will solve for this, presently unknown
m: total mass of the cart + bricks = 28.1 kg
k: coefficient of kinetic friction = 0.735
a: angle of rope with respect to the horizontal = 23 deg
L: distance moved = 19.8 m
g: acceleration due to gravity = 9.8 m/s2
Since the cart is moving at a constant speed, the net horizontal force on the cart should be 0, i.e., the horizontal pulling force from the rope should be exactly balanced out by frictional resistive force.
The frictional force (Coloumb friction) is given by (coeff of fricn)*(reaction force). In this case, the reaction force with the ground is not the weight (m*g) but, the weight - vertical lifting force from the rope. Therefore, the reaction force from the ground is, m*g-F*sin(a). In turn, the frictional resistive force is, k*(m*g-F*sin(a))
The horizontal pulling force of the rope is F*cos(a). Using our deduction that the horizontal pulling force must equal the frictional resistive force, we can solve for F.
F*cos(a) = k*(m*g - F*sin(a)) -----------(Eq 1)
Hence, F = k*m*g/(cos(a) + k*sin(a))
The total mechanical work done on an object by external forces is given by the dot product of the the net force vector with the net displacement vector.
Hence, once we know F, then the total work done is easily found because, mechanical work done in this case is (horizontal pulling force)*(horizontal distance travelled). Therefore,
Mechanical Work = F*cos(a)*L -----------(Eq 2)
where, F is solved using (Eq 1) and L, the displacement and and a the angle are known quantities.
So, mechanical work = k*m*g*L*cos(a)/(cos(a)+k*sin(a))
plugging in the values, the final answer would be, work = 3054.6 Joules.
2006-07-13 13:09:23 · answer #1 · answered by v_madhu 2 · 4⤊ 0⤋
The rope doesn't do any work on the cart becase W = F*d.
The rope exerts a force but it doesn't move in relation to the cart! d=0.
The total work expended (by whats pulling the cart) would be W = F*d. F = component in direction of motion = F_f * cos(23º), where F_f is the force due to friction .
F_f = u*n, where u is the coefficient of friction, n is the normal force.
u = .735, n = 28.1 * 9.81
Putting it all together,
W = F * d = .735 * 28.1 * 9.81 * cos(23º) * 19.8
2006-07-13 12:16:56 · answer #2 · answered by j 2 · 0⤊ 1⤋
Force of friction can be calculated by:
F(friction) = coefficient x (m x g - F x sin23)
F(friction) = 0.735 x (28.1 x 9.8 - 0.39 x F)
F(friction) = 202.404 - 0.287 x F
Because the block move at steady speed, and there are no acceleration so
F(friction) = F(in foward direction)
F(friction) = F x cos23
202.404 - 0.287 x F = 0.920 x F
202.404 = 1.207 x F
F = 167.692 Newton
So the work done is
W = F cos23 x S
W = 167.692 x 0.902 x 19.8
W = 3056.35 Joule
2006-07-13 17:44:55 · answer #3 · answered by Anonymous · 1⤊ 0⤋
work = force * distance
Force will be the force that is pulling the cart so that there is no acceleration. That means the force has to be equal to the frictional force.
Frictional force = coefficient * normal force.
Normal force = force perpendicular to the surface of the cart. (use your free body diagrams to figure it out)
Once you have Normal force you should get fricitional force which will give you the force needed to pull it across. Once you have that force, you take the distance and use the formula to get the work.
2006-07-13 12:06:29 · answer #4 · answered by Vicente 6 · 0⤊ 1⤋
Could this be a trick question? I can see asking how much force the rope exerts on the cart, but ropes can't actually do work. They don't expend energy. If the question asked about a horse, that would be different. So, I'll say the answer is no work whatsover.
2006-07-13 12:06:16 · answer #5 · answered by arbeit 4 · 0⤊ 1⤋
I'm not familiar with the term kinetic friction, but I hope this can help.
The work is: rope force*cos 23 * 19.8
first get the rope force from:
rope force*cos 23=(28.1*9.81-rope force*sin 23)*0.735
2006-07-13 12:22:22 · answer #6 · answered by Anonymous · 0⤊ 1⤋
i imagine being 9 months pregnant it doesn't be an truly wise concept for me to attempt a cart wheel. in spite of the indisputable fact that, my toddler actual looks in a position to do cart wheels in my abdomen!
2016-12-01 05:57:28 · answer #7 · answered by Anonymous · 0⤊ 0⤋
use mewk=Fn/Fk
you get 38.23129N as friction
since it was pulled at a constant speed, friction equals the force.
then you use W=Fd
W=(38.23129N)(19.1m)
W=730.2177N
please tell me if i'm wrong
2006-07-13 14:03:38 · answer #8 · answered by Rajan 3 · 0⤊ 1⤋