A toboggan, with rider, has a combined mass of 60kg at the top of a 5.0m hill. They are given an initial push of 2.0m/s. The distance down the hill was 20m. The velocity at the bottom of the hill is 10m/s. If the actual speed at the bottom was 9m/s, what would have been the average coefficient of friction between the toboggan and snow?
initial energy+PE added=final KE+frictionwork
1/2 *60*2^2+60*9.8*5=1/2 60*9^2+60*9.8*mu*cosineTheta
solve for mu, the coefficent of friction.
You actually have to know the slope of the hill.
tan theta=5/20=1/4
theta=arctan(.25)=14 degrees
cosTheta(14deg)= 0.97
I suspect you are assuming that g = 10 m/s^2
total energy at top = (1/2) m (2)^2 + m(2)(5) = 2 m + 50 m = 52 m Joules
energy at bottom at 10 m/s = (1/2) m (100) = 50 m Joules
If the speed at the bottom was only 9m/s
then
energy at bottom = (1/2) m (81) = 40.5 m Joules
52 - 40.5 = 11.5 m Joules lost to friction
Friction force * distance traveled = work done against friction
F * 20 = 11.5
F = .575 Newtons
that is
mu m g cos(slope angle) = .575
sin slope = 5/20
so
cos slope = .968
mu (60)(10)(.968) = .575
mu = 9.9 * 10^-4 = 0.00099
Friction force * distance traveled = work done against friction
F * 20 = 11.5 m
F = .575 m Newtons
that is
mu m g cos(slope angle) = .575 m
sin slope = 5/20
so
cos slope = .968
mu (60)(10)(.968) = .575 (60)
mu = .0594
(my previous answer times m, 60 kg
In other words the answer does not depend on the mass at all.
To find the average coefficient of friction between the toboggan and snow, we can use the concept of conservation of energy.
The total mechanical energy of the toboggan and rider at the top of the hill is equal to the total mechanical energy at the bottom of the hill, assuming no external forces like friction or air resistance are acting on the system.
The total mechanical energy (E) is given by:
E = KE + PE
Where KE is the kinetic energy and PE is the potential energy.
At the top of the hill, all the energy is in the form of potential energy, since the toboggan is stationary. So, the initial potential energy (PEi) is:
PEi = m * g * h
Where m is the mass of the toboggan and rider (60 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height of the hill (5.0 m).
PEi = 60 kg * 9.8 m/s^2 * 5.0 m = 2940 J
At the bottom of the hill, all the energy is in the form of kinetic energy, since the toboggan has reached its maximum speed. So, the final kinetic energy (KEf) is:
KEf = (1/2) * m * v^2
Where v is the velocity at the bottom of the hill (9 m/s).
KEf = (1/2) * 60 kg * (9 m/s)^2 = 2430 J
Since there is no energy loss due to external forces, the total mechanical energy is conserved:
PEi = KEf
2940 J = 2430 J
Now, let's consider the work done by friction along the 20 m distance down the hill. The work done by friction is given by:
Work = force of friction * distance
Since the work done by friction is equal to the change in mechanical energy, we can write:
Work = ΔE
Therefore, we have:
(friction force) * (20 m) = -510 J
Note that the negative sign indicates that work is being done against the direction of motion.
Now, let's calculate the friction force:
(friction force) = -(510 J) / (20 m) = -25.5 N
The negative sign indicates that the friction force acts in the opposite direction to the motion of the toboggan.
Finally, we can find the average coefficient of friction (µ) using the equation:
(friction force) = µ * (normal force)
In this case, the normal force is equal to the weight of the toboggan and rider:
(normal force) = m * g = 60 kg * 9.8 m/s^2 = 588 N
Therefore, the average coefficient of friction can be calculated as:
µ = (friction force) / (normal force) = (-25.5 N) / (588 N) = -0.0434
The average coefficient of friction between the toboggan and snow would be approximately 0.0434 (rounded to three decimal places).