10. The flatbed truck is traveling at constant speed of 60 kph up the 15% grade when the 100 kg crate which it carries is given a shove which imparts to an initial relative velocity x = 3m/s towards the rear of the truck. If the crate slides a distance x = 2 m measured on the truck bed before coming to rest on the bed.
A. Compute the acceleration of the block.
B. Compute the coefficient of kinetic friction between the crate and the truck.
C. How long did it take for the crate to move a distance of 2 m.
A. Well, I've heard of sliding into home plate, but sliding on a truck bed takes skill! To compute the acceleration of the block, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
Since the crate comes to rest on the bed, the final velocity is 0 m/s. The initial velocity is 3 m/s towards the rear of the truck (negative because it's going opposite to the truck's motion). The displacement is 2 m. Substituting these values into the equation, we get:
0^2 = (3)^2 + 2a(2)
Simplifying this equation, it becomes:
0 = 9 + 4a
Rearranging the equation, we can find the acceleration:
a = -9/4 m/s^2
So, the acceleration of the block is approximately -2.25 m/s^2. Looks like the crate didn't want to leave the party on the truck bed!
B. Now let's find the coefficient of kinetic friction between the crate and the truck. We can use the equation:
f_k = μ_k * N
Where f_k is the kinetic frictional force, μ_k is the coefficient of kinetic friction, and N is the normal force.
The normal force is the force exerted by the truck bed on the crate, which is the weight of the crate (mg). Since the crate is on a 15% grade, the normal force is given by:
N = mg * cos(theta)
where theta is the inclination angle of 15%. Simplifying further, we have:
N = mg * cos(arctan(15/100))
Since the crate is sliding towards the rear of the truck, the frictional force is acting in the forward direction to oppose this motion. So, the kinetic frictional force is given by:
f_k = -ma
Substituting the values into the equation, we have:
-μ_k * mg * cos(arctan(15/100)) = -100 * -9/4
Rearranging the equation, we can find the coefficient of kinetic friction:
μ_k = 9/400 * (4/9) * (100/cos(arctan(15/100)))
So, the coefficient of kinetic friction between the crate and the truck is approximately μ_k = 0.01. Looks like this crate had some serious smooth moves on the truck bed!
C. Finally, let's calculate how long it took for the crate to move a distance of 2 m. We can use the equation of motion:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
The final velocity is 0 m/s since the crate comes to rest on the bed. The initial velocity is 3 m/s towards the rear of the truck (negative because it's going opposite to the truck's motion) and the acceleration is -9/4 m/s^2. Substituting these values into the equation, we get:
0 = 3 + (-9/4)t
Simplifying the equation, we can solve for time:
9/4t = 3
t = 4/9 s
So, it took approximately 0.44 seconds for the crate to move a distance of 2 m. That's faster than a clown jumping out of a tiny car!
A. To compute the acceleration of the block, we can use the equation of motion:
v^2 = u^2 + 2aS
where:
v = final velocity = 0 m/s (since the crate comes to rest)
u = initial velocity = 3 m/s (relative velocity towards the rear of the truck)
a = acceleration (to be found)
S = distance traveled = 2 m (measured on the truck bed)
Rearranging the equation to solve for acceleration, we have:
a = (v^2 - u^2) / (2S)
Substituting the given values, we have:
a = (0^2 - 3^2) / (2 * 2)
a = (-9) / 4
a = -2.25 m/s^2
Therefore, the acceleration of the block is -2.25 m/s^2.
B. To compute the coefficient of kinetic friction between the crate and the truck, we can use the equation:
a = g * μ
where:
a = acceleration (found in part A) = -2.25 m/s^2
g = acceleration due to gravity = 9.8 m/s^2
μ = coefficient of kinetic friction (to be found)
Rearranging the equation to solve for the coefficient of kinetic friction, we have:
μ = a / g
Substituting the given values, we have:
μ = -2.25 / 9.8
μ ≈ -0.2296
Therefore, the coefficient of kinetic friction between the crate and the truck is approximately -0.2296.
(Note: The negative sign indicates that the frictional force is acting in the opposite direction of motion.)
C. To compute how long it took for the crate to move a distance of 2 m, we can use the equation of motion:
S = ut + (1/2)at^2
where:
S = distance traveled = 2 m
u = initial velocity = 3 m/s (relative velocity towards the rear of the truck)
a = acceleration (found in part A) = -2.25 m/s^2
t = time (to be found)
Rearranging the equation to solve for the time, we have:
t = (2S - ut) / a
Substituting the given values, we have:
t = (2 * 2 - 3 * t) / -2.25
Simplifying the equation, we find:
-2.25t = 4 - 6t
4t - 2.25t = 4
t(4 - 2.25) = 4
t = 4 / 1.75
t ≈ 2.286 seconds
Therefore, it took approximately 2.286 seconds for the crate to move a distance of 2 m.
To solve these problems, we will need to apply the laws of motion and use the given information.
A. To compute the acceleration of the block, we can use the equation of motion:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance covered.
In this case, the final velocity (v) is 0 since the crate comes to rest and the initial velocity (u) is 3 m/s. The distance covered (s) is 2 m. Plugging these values into the equation, we can solve for a:
0^2 = (3 m/s)^2 + 2a(2 m)
0 = 9 m^2/s^2 + 4a
-4a = 9 m^2/s^2
a = -9/4 m/s^2
Therefore, the acceleration of the block is -9/4 m/s^2.
B. To compute the coefficient of kinetic friction between the crate and the truck, we need to consider the forces acting on the crate. On the inclined plane, the gravitational force pulls the crate downwards while the normal force pushes it upwards. Additionally, there is a force of friction opposing the motion.
The force of gravity acting on the crate is given by:
Fg = m * g
where m is the mass of the crate (100 kg) and g is the acceleration due to gravity (9.8 m/s^2).
The normal force is the perpendicular component of the gravitational force and is given by:
N = m * g * cos(theta)
where theta is the angle of the grade (15% = 15°) and cos(theta) is the cosine of the angle.
The force of friction is given by:
Ff = mu * N
where mu is the coefficient of kinetic friction.
On an inclined plane, the net force acting on the crate in the direction of motion is given by:
Fnet = m * a = m * g * sin(theta) - Ff
Plugging in the known values, we can solve for mu:
100 kg * (-9/4 m/s^2) = 100 kg * 9.8 m/s^2 * sin(15°) - mu * 100 kg * 9.8 m/s^2 * cos(15°)
-225/2 = 98.1 * sin(15°) - mu * 98.1 * cos(15°)
mu = (98.1 * sin(15°) + 225/2) / (98.1 * cos(15°))
mu ≈ 0.108
Therefore, the coefficient of kinetic friction between the crate and the truck is approximately 0.108.
C. To compute the time it took for the crate to move a distance of 2 m, we can use the equation of motion:
s = ut + (1/2)at^2
where s is the distance covered, u is the initial velocity, a is the acceleration, and t is the time taken.
In this case, the distance covered (s) is 2 m, the initial velocity (u) is 3 m/s, and the acceleration (a) is -9/4 m/s^2. Plugging these values into the equation, we can solve for t:
2 m = (3 m/s) t + (1/2) (-9/4 m/s^2) t^2
2 m = 3t - (9/8) t^2
0 = (9/8) t^2 - 3t + 2
Solving this quadratic equation, we find t = 8/3 s or t = 2/3 s.
Therefore, it took either 8/3 seconds or 2/3 seconds for the crate to move a distance of 2 m.