jack gives 50 kg jill a ride on a sleigh for 500 meters. he exerts a force of 45 Newtons on the rope of the sled at an angle of 30 degrees. He pulls her at a constant velocity. What is the coefficient of friction between the sled and the ground? How much work does he do on the sled?
ok, im not good at physics but i'll give it a try. this may be a silly question but the ground they are on is snow right?
yes i guess it would be snow
The equation of motion is (in vector form)
vector (m•g) + vector F(fr)+ vector N+vectorF =0
Projections on the axes
x: F•cosα-F(fr) = 0
y: -mg+N+F•sinα = 0
F(fr) = F•cosα,
N =mg- F•sinα,
F(fr) = k•N =k•(mg- F•sinα),
F•cosα= k•(mg- F•sinα),
k = F•cosα/(mg- F•sinα) =
= 45•0.866/(50•9.8-45•0.5) = 0.083.
W= W(F) + W(fr)=F•cosα•s +F(fr) •s=
= F•cosα•s +k•N•s =
= s• (F•cosα + k•(mg - F•sinα)) =
=500 (45•0.866 + 0.083•(50•9.8 - 45•0.5)=38885 J
To find the coefficient of friction between the sled and the ground, we first need to determine the normal force acting on the sled.
The normal force is equal to the weight of the sled, which can be calculated using the formula: weight = mass × acceleration due to gravity.
Given that the mass of the sled is 50 kg, and the acceleration due to gravity is approximately 9.8 m/s^2, we can calculate the weight:
weight = 50 kg × 9.8 m/s^2 = 490 N
Since the sled is being pulled at a constant velocity, the force of friction is equal in magnitude and opposite in direction to the force Jack exerts on the rope.
We can break down the force Jack exerts on the rope into horizontal and vertical components. The vertical component (force perpendicular to the ground) doesn't affect the sled's motion because it cancels out the normal force.
The horizontal component (force parallel to the ground) is responsible for overcoming the force of friction.
The horizontal component of the force is given by: force_horizontal = force × cos(angle)
force_horizontal = 45 N × cos(30 degrees) ≈ 45 N × 0.866 ≈ 38.97 N
Since the sled is moving at a constant velocity, the force of friction is equal in magnitude and opposite in direction to the horizontal component of the force Jack exerts. Therefore, the force of friction is approximately 38.97 N.
Finally, we can calculate the coefficient of friction using the formula: coefficient of friction = force of friction / normal force.
coefficient of friction = 38.97 N / 490 N = 0.0798 (rounded to four decimal places)
Therefore, the coefficient of friction between the sled and the ground is approximately 0.0798.
Now, let's calculate the work done by Jack on the sled. Since the sled is moving at a constant velocity, there is no change in kinetic energy, so the work done is zero.
The work done on an object is equal to the force applied multiplied by the distance over which the force is applied: work = force × distance × cos(angle).
In this case, Jack exerts a force of 45 N, and the sled is pulled for a distance of 500 meters:
work = 45 N × 500 m × cos(30 degrees) ≈ 45 N × 500 m × 0.866 ≈ 19,688.25 J
Therefore, Jack does approximately 19,688.25 Joules of work on the sled.