If motor M exerts a force of
F=(10t2+100)N on the cable, where t is in seconds, determine the velocity of the
25kg crate when
t=4s . The coefficients of static and kinetic friction between the crate and the plane are
μs=0.3 and 0.25
μk=0.25 . The crate is initially at rest.
Well, this seems like quite the problem. But don't worry, I'm here to make it a little more entertaining!
First, let's find the net force acting on the crate. The force of motor M is given by F = 10t^2 + 100 N, so when t = 4s, we can calculate that F = 10(4^2) + 100 = 260 N.
Now, let's calculate the maximum static friction force. The coefficient of static friction is μs = 0.3, and the weight of the crate is given by W = mg = 25kg * 9.8m/s^2 = 245N. The maximum static friction force is µs * W = 0.3 * 245N = 73.5N.
Since the applied force F = 260N is greater than the maximum static friction force, the crate will start moving, and the static friction is no longer a factor.
Once the crate starts moving, the coefficient of kinetic friction comes into play. The coefficient of kinetic friction is μk = 0.25. So, the kinetic friction force is given by μk * W = 0.25 * 245N = 61.25N.
Now, we can calculate the net force on the crate as follows: Net Force = F - kinetic friction force = 260N - 61.25N = 198.75N.
Using Newton's second law, F_net = ma, where m is the mass of the crate (25kg), and a is the acceleration. We can rearrange the equation to solve for the acceleration: a = F_net / m = 198.75N / 25kg = 7.95m/s^2.
Finally, let's find the velocity of the crate when t = 4s. Since the crate starts from rest, we can use the kinematic equation v = u + at, where u is the initial velocity (0m/s) and t is the time (4s). Plugging in the values, we get v = 0m/s + 7.95m/s^2 * 4s = 31.8m/s.
So, the velocity of the 25kg crate when t = 4s is 31.8m/s. Now that's some fast-moving comedy!
To determine the velocity of the crate, we need to consider the net force acting on it.
First, let's calculate the force of friction acting on the crate.
The gravitational force acting on the crate can be calculated using the equation:
F_gravity = m * g
where m is the mass of the crate (25 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2).
F_gravity = 25 kg * 9.8 m/s^2
F_gravity = 245 N
The maximum static friction force can be calculated using the equation:
F_static_friction = μs * F_normal
where μs is the coefficient of static friction and F_normal is the normal force acting on the crate. In this case, the normal force is equal to the gravitational force.
F_static_friction = 0.3 * 245 N
F_static_friction = 73.5 N
Since the maximum static friction force is less than the applied force of 10t^2 + 100 N, the crate will not start moving. Therefore, the crate remains at rest when t = 4 s.
To confirm this, we can calculate the force of static friction at t = 4 s:
F_applied = 10t^2 + 100 N
F_applied = 10 * (4^2) + 100 N
F_applied = 160 + 100 N
F_applied = 260 N
The force of static friction (73.5 N) is less than the applied force (260 N). Since the force of static friction opposes the applied force, the crate will not start moving.
Therefore, the velocity of the crate when t = 4 s is 0 m/s.
To determine the velocity of the crate when t = 4s, we need to consider the forces acting on the crate and apply Newton's laws of motion.
First, let's analyze the forces acting on the crate. We have the force exerted by the motor, F = (10t^2 + 100) N, which we can substitute for t = 4s to get F = (10(4^2) + 100) N = 260 N.
Next, we have the force of friction. Since the crate is initially at rest, we need to calculate the static friction force before considering kinetic friction. The equation for static friction is given by Fs ≤ μsN, where μs is the coefficient of static friction and N is the normal force acting on the crate.
The normal force, N, is equal to the weight of the crate, which is given by N = mg, where m is the mass of the crate (25 kg) and g is the acceleration due to gravity (approximately 9.8 m/s^2). Hence, we have N = 25 kg × 9.8 m/s^2 = 245 N.
Substituting these values into the equation for static friction, we get Fs ≤ 0.3 × 245 N = 73.5 N. Since the crate is initially at rest, the static friction will exactly balance the force applied by the motor (260 N) until it reaches its maximum value of 73.5 N.
Therefore, when t = 4s, the static friction force acting on the crate by the plane is equal to 73.5 N.
Once the static friction force is exceeded, the crate will start moving, and the kinetic friction force will come into play. The equation for kinetic friction is given by Fk = μkN, where μk is the coefficient of kinetic friction and N is the normal force.
Substituting the values, we get Fk = 0.25 × 245 N = 61.25 N.
Since the force applied by the motor (260 N) exceeds the maximum static friction (73.5 N), the crate will start moving, and the kinetic friction force will act on it, opposing its motion.
To determine the crate's acceleration, we can use Newton's second law: Fnet = ma, where Fnet is the net force acting on the crate and a is its acceleration.
The net force is given by Fnet = F - Fk, where F is the force applied by the motor (260 N) and Fk is the kinetic friction force (61.25 N). Substituting these values, we have Fnet = 260 N - 61.25 N = 198.75 N.
Using Newton's second law, we have Fnet = ma, where m is the mass of the crate (25 kg) and a is the acceleration. Rearranging the equation, we get a = Fnet/m = 198.75 N/25 kg = 7.95 m/s^2.
Now that we have the acceleration of the crate, we can determine its velocity at t = 4s using the kinematic equation: v = u + at, where v is the final velocity, u is the initial velocity (which is 0 since the crate is initially at rest), a is the acceleration, and t is the time.
Substituting the values, we get v = 0 + (7.95 m/s^2) × (4 s) = 31.8 m/s.
Therefore, when t = 4s, the velocity of the 25kg crate is 31.8 m/s.
F = 10*4^2 + 100 = 260 N.
Wc = m*g = 25 * 9.8 = 245 N. = Wt. of crate.
Fk = uk*mg = 0.25 * 245 = 61.25 N. =
Force of kinetic friction.
a = Fn/m = (F-Fk)/m = (260-61.25)/25 =
7.95 m/s^2.
V = a * t = 7.95 * 4 = 31.8 m/s.