The coefficients of static and kinetic friction between a 50 kg box and a horizontal surface are 0.64 and 0.17, respectively.
(a) What is the acceleration of the object if a 335 N horizontal force is applied to the box?
m/s2
(b) What is the acceleration if the applied force is 465 N
(a) Well, well, well, looks like our box is going on a little ride! To calculate the acceleration, we need to consider the forces at play. We have our applied force, opposing the force of friction. The coefficient of static friction is 0.64, so we'll need to multiply that by the normal force (which is equal to the weight of the object, given that it's on a horizontal surface). Hang in there, we're getting to the fun part!
First, let's find the force of static friction:
Fs = μs * N
= 0.64 * (50 kg * 9.8 m/s^2)
= 313.6 N
(Note: N represents the normal force, which equals the weight of the object)
Now, since the applied force (Fa) is greater than the force of static friction (Fs), the box starts moving. At this point, we're dealing with kinetic friction. The force of kinetic friction is given by:
Fk = μk * N
= 0.17 * (50 kg * 9.8 m/s^2)
= 83.3 N
To find the acceleration, we subtract the force of kinetic friction from the applied force and divide by the mass of the box:
a = (Fa - Fk) / m
= (335 N - 83.3 N) / 50 kg
= 5.03 m/s^2
So, the acceleration of the object with a 335 N force is 5.03 m/s^2. Hold on tight!
(b) Alright, let's rev up the acceleration with a 465 N force. We follow the same steps as before. The force of static friction remains the same, 313.6 N (0.64 * 50 kg * 9.8 m/s^2), but the force of kinetic friction increases to:
Fk = 0.17 * (50 kg * 9.8 m/s^2)
= 83.3 N
Now, we calculate the acceleration with the new applied force:
a = (Fa - Fk) / m
= (465 N - 83.3 N) / 50 kg
= 7.63 m/s^2
So, with a 465 N force, the acceleration turns up to 7.63 m/s^2. Hang on tight, things are getting faster and funnier!
To calculate the acceleration of the object given the applied force and the coefficients of static and kinetic friction, we need to consider two scenarios: when the object is at rest (static friction) and when the object is in motion (kinetic friction).
(a) When a 335 N horizontal force is applied to the box:
To determine whether the object will start moving or remain at rest, we compare the applied force to the maximum static friction force.
The maximum static friction force (fs_max) can be calculated by multiplying the coefficient of static friction (μs) by the normal force (N), where N equals the weight of the object (mg).
Given:
Mass of the box (m) = 50 kg
Coefficient of static friction (μs) = 0.64
Coefficient of kinetic friction (μk) = 0.17
Applied force (F) = 335 N
Weight of the box (mg) = mass x acceleration due to gravity
= 50 kg x 9.8 m/s^2
= 490 N
Maximum static friction force (fs_max) = μs * N
= 0.64 * 490 N
= 313.6 N
Since the applied force (335 N) is greater than the maximum static friction force (313.6 N), the object will overcome static friction and start moving.
Therefore, the acceleration of the object will be the same as the kinetic friction force (fk), divided by the mass (m).
Kinetic friction force (fk) = μk * N
= 0.17 * 490 N
= 83.3 N
Acceleration (a) = fk / m
= 83.3 N / 50 kg
= 1.67 m/s^2
Therefore, the acceleration of the object is 1.67 m/s^2.
(b) When the applied force is 465 N:
We need to check whether the applied force is greater than the maximum static friction force (313.6 N) to determine if the object will start moving or remain at rest.
Since the applied force (465 N) is greater than the maximum static friction force (313.6 N), the object will overcome static friction and start moving.
Thus, we use the same calculation for the acceleration as in part (a), using the coefficient of kinetic friction (μk) instead of the coefficient of static friction.
Kinetic friction force (fk) = μk * N
= 0.17 * 490 N
= 83.3 N
Acceleration (a) = fk / m
= 83.3 N / 50 kg
= 1.67 m/s^2
Therefore, the acceleration of the object, when the applied force is 465 N, is also 1.67 m/s^2.
To find the acceleration of the object, we need to consider both the forces acting on it and the frictional forces.
(a) To calculate the acceleration when a 335 N force is applied:
1. Calculate the maximum force of static friction using the equation: fs = µs * N, where µs is the coefficient of static friction and N is the normal force.
fs = 0.64 * (50 kg * 9.8 m/s^2) = 313.6 N
2. Determine if the applied force is greater or smaller than the maximum force of static friction:
If the applied force is less than the maximum static friction force, the object will not move and the acceleration is 0 m/s^2.
If the applied force is equal to or greater than the maximum static friction force, the object will start moving, and we need to consider kinetic friction.
3. Calculate the force of kinetic friction using the equation: fk = µk * N, where µk is the coefficient of kinetic friction.
fk = 0.17 * (50 kg * 9.8 m/s^2) = 83.3 N
4. Subtract the force of kinetic friction from the applied force to find the net force: Fnet = Fapplied - fk
Fnet = 335 N - 83.3 N = 251.7 N
5. Finally, use Newton's second law of motion, F = ma, to calculate the acceleration:
251.7 N = (50 kg) * a
a = 251.7 N / 50 kg = 5.03 m/s^2
Therefore, the acceleration of the object is 5.03 m/s^2.
(b) To calculate the acceleration when a 465 N force is applied:
Follow steps 1-4 from part (a):
1. Calculate the maximum force of static friction:
fs = 0.64 * (50 kg * 9.8 m/s^2) = 313.6 N
2. Determine if the applied force is greater or smaller than the maximum force of static friction:
Since the applied force of 465 N is greater than the maximum static friction force, the object will start moving, and we need to consider kinetic friction.
3. Calculate the force of kinetic friction:
fk = 0.17 * (50 kg * 9.8 m/s^2) = 83.3 N
4. Subtract the force of kinetic friction from the applied force to find the net force:
Fnet = 465 N - 83.3 N = 381.7 N
5. Use Newton's second law of motion to calculate the acceleration:
381.7 N = (50 kg) * a
a = 381.7 N / 50 kg = 7.63 m/s^2
Therefore, the acceleration of the object when a 465 N force is applied is 7.63 m/s^2.
(a) It takes a force of 0.64 M g = 314 N to make it move, so it will move in this case. The kinetic friction force opposing motion will be
0.17 M g = 83 N. The net applied force is
Fnet = 335 - 83 = 252 N
Acceleration = Fnet/M = 5.04 m/s^2
(b) The method of solution is the same.