In a pickup game of dorm shuffleboard, students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.88 m by the horizontal 25 N force from the broom and then has a speed of 1.63 m/s, what is the coefficient of kinetic friction between the book and floor?
Work done by broom = Kinetic energy increase + work done against fraction
Compute the friction work using that equation. Divide it by the distance moved to get the friction force. Use that to get the friction coefficient
To find the coefficient of kinetic friction between the book and the floor, we can use Newton's laws of motion.
First, let's calculate the net force acting on the book. The net force can be calculated using Newton's second law:
Net force (F_net) = mass (m) × acceleration (a)
The book's acceleration can be found using the equation:
v^2 = u^2 + 2as
where:
v = final velocity (1.63 m/s)
u = initial velocity (0 m/s)
a = acceleration
s = distance (0.88 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Substituting the values, we have:
a = (1.63^2 - 0^2) / (2 × 0.88)
Next, we can calculate the net force:
F_net = m × a
F_net = 3.5 kg × [(1.63^2 - 0^2) / (2 × 0.88)]
Now, let's consider the forces acting on the book. There are two forces:
1. The force applied by the broom, which is 25 N.
2. The force due to kinetic friction between the book and the floor, which can be calculated as:
Frictional force (F_friction) = coefficient of kinetic friction (μ) × normal force (F_normal)
where the normal force is equal to the weight of the book:
F_normal = m × g
where:
m = mass (3.5 kg)
g = acceleration due to gravity (9.8 m/s^2)
Let's calculate the normal force:
F_normal = 3.5 kg × 9.8 m/s^2
Finally, the net force can be written as:
F_net = F_applied - F_friction
Substituting the values, we have:
3.5 kg × [(1.63^2 - 0^2) / (2 × 0.88)] = 25 N - (μ × 3.5 kg × 9.8 m/s^2)
Now, let's solve for the coefficient of kinetic friction (μ). Rearranging the equation, we have:
μ = (25 N - 3.5 kg × [(1.63^2 - 0^2) / (2 × 0.88)]) / (3.5 kg × 9.8 m/s^2)
Calculating the expression above will give us the coefficient of kinetic friction between the book and the floor.
To find the coefficient of kinetic friction between the book and the floor, we need to use the principles of Newton's second law and the equation for kinetic friction.
1. First, let's calculate the net force acting on the book:
Net force = Applied force - Frictional force
2. The applied force is given as 25 N in the horizontal direction.
3. The frictional force can be calculated using the equation for kinetic friction:
Frictional force = coefficient of kinetic friction * Normal force
4. The normal force is the force exerted by the floor perpendicular to the book's surface. Since the book is on a horizontal surface, the normal force cancels out the gravitational force acting on the book:
Normal force = gravitational force = mass * gravity
5. The mass of the book is given as 3.5 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
Now, let's calculate the frictional force and the coefficient of kinetic friction:
Step 1: Calculate the normal force:
Normal force = mass * gravity = 3.5 kg * 9.8 m/s^2
Step 2: Calculate the frictional force:
Frictional force = coefficient of kinetic friction * Normal force
Step 3: Calculate the net force:
Net force = Applied force - Frictional force
Step 4: Use Newton's second law to find the acceleration:
Net force = mass * acceleration
Step 5: Use the distance and final velocity to find the acceleration:
v^2 = u^2 + 2as
Final velocity (v) = 1.63 m/s
Initial velocity (u) = 0 m/s (starting from rest)
Distance (s) = 0.88 m
Step 6: Use the acceleration to find the coefficient of kinetic friction:
coefficient of kinetic friction = frictional force / (mass * acceleration)
Plug in the values and follow these steps to calculate the coefficient of kinetic friction between the book and the floor.
m = 3.5kg
x = 0.88m
f = 25N
speed = 1.63m/s
coefficient of kinetic friction = ?
we know, fk=fnormal*coefficient of kinetic friction
where fk = kinetic frictional force
so, fk= mg*coefficient
fk=3.5*9.8*cefficient
a=?
so applying the equation
v^2=u^2+2ax
a= 1.1m/s^2
fnet=ma
so, ma=3.5*9.8*coefficient
cefficient= 0.11