A crate is given an initial speed of 3.1 m/s up the 29 degree plane shown in the figure . Assume coefficient of friction = 0.17.
How far up the plane will it go?
How much time elapses before it returns to its starting point?
To determine how far up the plane the crate will go, we can use the concept of work and energy. The work done on the crate equals the change in its kinetic energy.
Work done on the crate = Change in kinetic energy
Since the crate starts with an initial speed and comes to rest at the highest point, the change in kinetic energy is equal to its initial kinetic energy.
Initial kinetic energy = 0.5 * mass * (initial velocity)^2
Given that the initial speed of the crate is 3.1 m/s, we need to find the mass of the crate to calculate its initial kinetic energy. However, the mass of the crate is not provided in the question.
To determine the distance up the plane the crate will go, we can use the concept of projectile motion and trigonometry.
Distance up the plane = (Initial velocity)^2 * sin^2(angle) / (2 * acceleration)
The acceleration can be calculated using the equation:
Acceleration = g * sin(angle) - coefficient of friction * g * cos(angle)
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now let's compute the distance up the plane and the time it takes for the crate to return to its starting point when given the crate's mass.
To determine how far up the plane the crate will go, we can use the concept of work and energy. The work done on the crate against gravity and friction will be equal to the change in its kinetic energy.
First, let's calculate the work done against gravity. The gravitational force acting on the crate can be calculated using the equation F_gravity = mg, where m is the mass of the crate and g is the acceleration due to gravity (approximately 9.8 m/s^2). The component of this force parallel to the incline will be F_gravity_parallel = m * g * sin(θ), where θ is the angle of the plane.
Next, let's calculate the work done against friction. The frictional force can be calculated using the equation F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force acting on the crate. The normal force can be calculated as F_normal = m * g * cos(θ), since the crate is on an inclined plane.
The work done against friction will be W_friction = F_friction * d, where d is the distance traveled.
Since the crate starts from rest, its initial kinetic energy is 0. Therefore, the total work done on the crate will be equal to the negative of the work done against gravity and friction.
Now, let's calculate the work done on the crate:
W_total = - (W_gravity + W_friction)
= - (F_gravity_parallel * d + F_friction * d)
= - (m * g * sin(θ) * d + μ * m * g * cos(θ) * d)
Since work can also be expressed as the change in kinetic energy, we can set W_total equal to the change in kinetic energy:
W_total = ΔKE
Where ΔKE = KE_final - KE_initial.
Since the crate comes to rest at its highest point, the final kinetic energy is 0.
Therefore, we have:
- (m * g * sin(θ) * d + μ * m * g * cos(θ) * d) = 0
Simplifying the equation, we get:
d = -μ * tan(θ)
Substituting the known values:
d = -0.17 * tan(29°)
Calculating this, the distance up the plane will be approximately 0.16 meters.
To determine how much time elapses before the crate returns to its starting point, we can use the concept of time and distance. The time taken to reach the highest point of the motion will be half of the total time taken for a complete round trip.
To calculate the time taken for a round trip, we need to find the total distance traveled. Since the crate started from rest and returns to its starting point, the total distance will be twice the distance up the plane.
Total distance = 2 * distance up the plane
Substituting the known value:
Total distance = 2 * 0.16 meters
Calculating this, the total distance traveled will be approximately 0.32 meters.
Next, we can calculate the total time taken for the round trip using the equation:
Total time = (2 * distance) / initial velocity
Substituting the known values:
Total time = (2 * 0.32 meters) / 3.1 m/s
Calculating this, the total time taken for the round trip will be approximately 0.207 seconds.
Finally, to find the time taken before it returns to its starting point, we divide the total time by 2:
Time elapsed before it returns to the starting point = total time / 2
Substituting the known value:
Time elapsed before it returns to the starting point = 0.207 seconds / 2
Calculating this, the time elapsed before it returns to its starting point will be approximately 0.1035 seconds.
KE= W(fr) +PE
KE =m•v²/2
W(fr) =F(fr) •s = μ•m•g•s •cosα
PE=m•g•h = m•g•s•sinα
m•v²/2 = μ•m•g •s •cosα + m•g•s•sinα =
=m•g•s(μ•cosα + sinα).
s= v²/2•g•(μ•cosα + sinα)=
=3.1²/2•9.8•(0.17•0.875+0.48)=0.78 m.
a1 = v²/2•s = 3.1²/2•0.78 =6.16 m/s².
0=v –a•t1
t1=v/a=3.1/6.16 = 0.5 s. (the time of upward motion)
m•a=m•g•sinα - μ•m•g • cosα
a2= g(sinα - μ•cosα)=9.8•(sin29-0.17•cos29)=3.25 m/s².
s=a2•t2²/2,
t2=sqrt(2s/a2) = sqrt(2•0.78/3.25)=0.69 s
t=0.5+0.69=1.19 s.