Two crates of fruit are released from the top of a ramp inclined at 25 degrees from the horizontal and 2.5 meter long. The two crates consist of an apple crate of mass 35 kg that is placed in front of a watermelon crate of mass 80 kg. The apple crate has a coefficient of friction of 0.20 while the watermelon crate has a coefficient of friction of 0.15. How long does it take the apple crate to reach the bottom of the incline if it needs to travel a distance of 2.5 meters?
first find the retarding force on each.
watermelon: 80*.15*9.8*cos25deg=106.6N
apple: 35*.2*9.8cos25deg=62.2N
then, find the accelerating force on each.
watermellion=80*9.8*sin35deg-62.2= 387.5
apple: 35*9.8*sin35-62.2=134N
then we have to test if the apple goes fastest:
Accapplie=134/35=3.82 m/s^2
Awater=388/80=4.85 m/s^2
so, the watermellon is pushing the apple...
for the system joined,
acceleration=netforce/total mass=(388+134)/(115)=4.54m/s^2
time...
d=1/2 a t^2 or t= sqrt(2ad) you know a, and d.
To find out how long it takes the apple crate to reach the bottom of the incline, we need to determine the net force acting on the crate and then use Newton's second law of motion to calculate the acceleration. Finally, we can use the equation of motion to find the time taken.
Let's break down the problem step by step:
Step 1: Determine the gravitational force acting on each crate.
The gravitational force can be calculated using the equation F = m * g, where F is the force, m is the mass, and g is the acceleration due to gravity (approximately 9.8 m/s^2). For the apple crate, the force is 35 kg * 9.8 m/s^2 = 343 N. Similarly, for the watermelon crate, the force is 80 kg * 9.8 m/s^2 = 784 N.
Step 2: Determine the frictional force acting on each crate.
To calculate the frictional force, we use the equation F_friction = μ * F_normal, where μ is the coefficient of friction and F_normal is the normal force (equal to the component of the weight perpendicular to the incline). The normal force can be calculated as F_normal = m * g * cosθ, where θ is the angle of the ramp from the horizontal (25 degrees).
For the apple crate, the normal force is 35 kg * 9.8 m/s^2 * cos(25°) = 308.4 N. Thus, the frictional force on the apple crate is 0.20 * 308.4 N = 61.68 N.
Similarly, for the watermelon crate, the normal force is 80 kg * 9.8 m/s^2 * cos(25°) = 770.96 N. The frictional force on the watermelon crate is 0.15 * 770.96 N = 115.64 N.
Step 3: Determine the net force acting on the apple crate.
The net force is the difference between the gravitational force and the frictional force:
Net force on apple crate = 343 N - 61.68 N = 281.32 N.
Step 4: Calculate the acceleration of the apple crate.
Using Newton's second law of motion, F_net = m * a, where F_net is the net force, m is the mass, and a is the acceleration. Rearranging the formula, we have a = F_net / m.
Since the mass of the apple crate is 35 kg and the net force is 281.32 N, the acceleration of the apple crate is 281.32 N / 35 kg = 8.04 m/s^2.
Step 5: Calculate the time taken for the apple crate to reach the bottom.
We can use the equation of motion, s = u * t + (1/2) * a * t^2, where s is the distance, u is the initial velocity (which we assume to be zero), a is the acceleration, and t is the time taken.
Rearranging the equation, we have t^2 = (2 * s) / a. Plugging in the values, t^2 = (2 * 2.5 m) / 8.04 m/s^2 = 0.6211 s^2.
Therefore, the time taken for the apple crate to reach the bottom of the incline is t = √0.6211 s^2, which is approximately 0.788 seconds.
So, it takes approximately 0.788 seconds for the apple crate to reach the bottom of the incline.