A 16kg sled starts up a 28 degree incline with a speed of 2.4m/s . The coefficient of kinetic friction is = 0.27.
part a)How far up the incline does the sled travel?
part b)What condition must you put on the coefficient of static friction if the sled is not to get stuck at the point determined in part (a)?
KE =m•v²/2,
PE =m•g•h = m•g •s•sinα,
W(fr) = F(fr) •s = k•m•g•cosα •s
KE = PE +W(fr)
v²/2 = g •s(sinα + k•cosα),
s = v²/2•s(sinα + k•cosα),
KE =m•v²/2,
PE =m•g•h = m•g •s•sinα,
W(fr) = F(fr) •s = k•m•g•cosα •s
KE = PE +W(fr)
v²/2 = g •s(sinα + k•cosα),
s = v²/2•g•(sinα + k•cosα),
To find the distance the sled travels up the incline, we can use the concept of work done against friction. Work done is the force applied multiplied by the distance traveled.
First, let's find the force of friction acting on the sled. The frictional force can be calculated using the coefficient of kinetic friction and the normal force.
Frictional force = coefficient of kinetic friction * normal force.
The normal force can be calculated as the perpendicular component of the weight of the sled. It can be found using:
Normal force = mass * acceleration due to gravity * cos(angle of incline).
Substituting the given values:
Mass = 16 kg,
Acceleration due to gravity = 9.8 m/s^2,
Angle of incline = 28 degrees,
Coefficient of kinetic friction = 0.27.
Normal force = 16 kg * 9.8 m/s^2 * cos(28 degrees).
Next, let's calculate the frictional force:
Frictional force = 0.27 * (16 kg * 9.8 m/s^2 * cos(28 degrees)).
Now we can find the work done against friction. The work done against friction is equal to the force of friction multiplied by the distance traveled.
Let's assume the distance traveled up the incline is d.
Work done against friction = Frictional force * d.
Since the work done against friction is equal to the change in kinetic energy of the sled using the work-energy principle, we can equate the two.
Work done against friction = Change in kinetic energy.
Change in kinetic energy = (1/2) * mass * (final velocity)^2 - (1/2) * mass * (initial velocity)^2.
The initial velocity is given as 2.4 m/s, and the final velocity is assumed to be zero (as the sled should come to a stop at the top).
Now we can solve for the distance traveled up the incline, d, in part (a).
To answer part (b), the coefficient of static friction must be such that it is greater than or equal to the coefficient of kinetic friction. This ensures that the sled does not get stuck and starts moving up the incline. Therefore, the condition is:
Coefficient of static friction ≥ Coefficient of kinetic friction.
In this case, the coefficient of static friction is not explicitly given, but as long as it is greater than or equal to 0.27 (the coefficient of kinetic friction), the sled will not get stuck.