A horse pulls a child on a sled along a horizontal snow-covered trail using a rope. The horse does 8.0x10^3J of work as it pulls the child and sled 62m at a constant speed. The coefficient of friction between the sled and the snow is 0.25.
A) Find the tension in the rope
B) Find the combined mass of the child and the sled
W(horse) =W(fr) =μNs= μmgs
m=W/ μ gs= 8000/0.25•9.8•62=52.7 kg
If the rope is horizontal and
v=const =>
T= F(fr) = μmg=0.25•52.7•9.8 =129 N
Does your textbook give the mass of the sled?
Thanks ^^
To find the tension in the rope, we can use the work-energy principle. According to this principle, the work done on an object is equal to its change in kinetic energy. In this case, the work done by the horse is equal to the change in kinetic energy of the child and the sled.
A) Find the tension in the rope:
The work done by the horse is given as 8.0x10^3 J. This is equal to the change in kinetic energy (KE) of the child and the sled.
The change in kinetic energy (ΔKE) is given by:
ΔKE = KE_final - KE_initial
Since the child and sled start from rest (initial kinetic energy is zero), the change in kinetic energy is equal to the final kinetic energy:
ΔKE = KE_final = (1/2)mv^2
Where m is the combined mass of the child and the sled, and v is their constant speed.
Substituting the given values, we have:
8.0x10^3 J = (1/2)m(0)^2
8.0x10^3 J = 0
This equation doesn't provide any useful information, because it suggests that there is no work done and no change in kinetic energy. However, this is not possible since the horse is doing work to pull the child and sled.
To reconcile this, we can consider the work done against friction. The work done against friction is equal to the product of the frictional force and the distance over which it is applied. In this case, the frictional force can be calculated using the coefficient of friction (μ) and the normal force (N).
The frictional force (f) is given by:
f = μN
The normal force (N) is equal to the weight of the child and the sled (mg).
Substituting the given values, we have:
f = μmg
The work done against friction is then given by:
Work_friction = f × d
Where d is the distance over which the frictional force is applied, which is equal to the distance the child and sled move.
Now, we can go back to the work-energy principle equation:
8.0x10^3 J = ΔKE + Work_friction
Since ΔKE is still unknown, let's solve for Work_friction first.
Work_friction = f × d = (μmg) × d
Now, we have:
8.0x10^3 J = ΔKE + (μmg) × d
Since we know that ΔKE = KE_final, and the child and sled are moving at a constant speed, the final kinetic energy is also given by:
KE_final = (1/2)m(0)^2 = 0
Therefore, we have:
8.0x10^3 J = 0 + (μmg) × d
Now, we can solve for the tension in the rope. The tension in the rope (T) is equal to the work done by the horse plus the work done against friction. Since both work contributions are in opposite directions, we have:
T = 8.0x10^3 J + (μmg) × d
Substituting the given values for the coefficient of friction, mass, and distance, we can calculate the tension in the rope.