A 120-kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 400 N. For the first 17 m the floor is frictionless, and for the next 17 m the coefficient of friction is 0.34. What is the final speed of the crate?
find the work that went into the system: Force*distance
Now, find the frictional force on the next portion:
friction=mu*mg
Then find the work done on friction:
mu*mg*17
Finally, find the KE of the crate
KE crate=workin-frictionwork
1/2 mv^2=400*34-mu(mg)17
v^2=800*34/mass -2*mu*g*17
solve for v
check my thinking.
To find the final speed of the crate, we need to analyze the forces acting on it and calculate the net force and acceleration.
First, let's consider the motion of the crate on the frictionless floor. In this case, there is no friction force opposing the motion, so the only force acting on the crate is the applied force of 400 N. Since there are no other forces, the crate accelerates uniformly according to Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration (F=ma).
The mass of the crate is given as 120 kg, and the applied force is 400 N. So, we can calculate the acceleration using the formula:
acceleration = force / mass
acceleration = 400 N / 120 kg
Next, let's calculate the acceleration:
acceleration = 3.33 m/s^2
Now, we have determined the acceleration on the frictionless stretch of the floor. To find the final speed of the crate after covering 17 m on the frictionless floor, we can use the kinematic equation:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance
Since the crate starts from rest (initial velocity is zero), the equation simplifies to:
final velocity^2 = 2 * acceleration * distance
Plugging in the values, we get:
final velocity^2 = 2 * 3.33 m/s^2 * 17 m
final velocity^2 = 113.06 m^2/s^2
Taking the square root of both sides to find the final velocity, we get:
final velocity = √(113.06) m/s
final velocity ≈ 10.63 m/s
So, the final speed of the crate after covering the frictionless stretch is approximately 10.63 m/s.
Now, let's calculate the frictional force on the crate when it encounters the second, 17 m stretch of the floor with a coefficient of friction of 0.34. We can use the formula:
frictional force = coefficient of friction * normal force
The normal force is equal to the weight of the crate, which can be calculated as:
normal force = mass * gravity
where gravity is approximately 9.8 m/s^2.
normal force = 120 kg * 9.8 m/s^2
normal force ≈ 1176 N
Now, we can calculate the frictional force:
frictional force = 0.34 * 1176 N
frictional force ≈ 399.84 N
The frictional force is opposing the motion, so the net force acting on the crate is the difference between the applied force and the frictional force:
net force = applied force - frictional force
net force = 400 N - 399.84 N
net force ≈ 0.16 N
Since the net force is so small, the crate will not accelerate significantly while moving on the portion with friction. As a result, the final velocity on this portion will be close to the final velocity achieved on the frictionless portion.
Therefore, the final speed of the crate, after covering the entire distance of 34 m, is approximately 10.63 m/s.