Can't figure this one out.
"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?"
I know these values, but I just can't find a formula that uses velocity and distance to plug them into.
m = 3.5 kg
g = 9.8 m/s/s
d = 0.88 m
F = 25N
Fn = 34.3N (Fn = mg)
v = 1.63 m/s
a = 7.14 m/s/s (a = F/m)
Muk = ??
I'm not sure if I even need all of those values...can anyone help me out with this?
Subtract the final kinetic energy from the work done pushing the book, 25*0.88 joules
The difference will be work done against friction, which is Ff*X
Ff is the friction force, M*g*Muk
Solve for Muk
I used my last attempt at the problem using this info, but was unable to figure out the answer. Thank you for responding though.
W = 22 J = total work done
Final KE = (1/2)MV^2 = 4.65 J
Friction work done = 22-4.65 = 17.35 J
Friction force = 17.35/X = 19.7 N
Muk = 19.7/(M*g) = 0.575
(Muk is a dimensionless ratio)
To solve this problem, you need to use the concept of work and energy.
First, let's calculate the work done by the broom when it pushes the book. Work is defined as the force applied multiplied by the distance over which it acts. In this case, the force is 25 N and the distance is 0.88 m. So the work done by the broom is:
Work = Force x Distance
Work = 25 N x 0.88 m
Work = 22 J
Next, let's consider the energy of the book. Initially, the book is at rest, so its initial kinetic energy is zero. Finally, when the book reaches a speed of 1.63 m/s, it has some final kinetic energy. The change in kinetic energy of the book is equal to the work done on it (according to the work-energy theorem). So we can write:
Change in Kinetic Energy = Work done on the book
The final kinetic energy is given by the formula:
Final Kinetic Energy = (1/2) x mass x velocity^2
Since the book has a mass of 3.5 kg and a velocity of 1.63 m/s, we can substitute these values into the formula:
Final Kinetic Energy = (1/2) x 3.5 kg x (1.63 m/s)^2
Final Kinetic Energy = 6.2975 J
Now, we can equate the change in kinetic energy to the work done on the book:
Change in Kinetic Energy = Work done on the book
6.2975 J = 22 J
Since the change in kinetic energy is less than the work done, we can conclude that some of the work done is used to overcome friction between the book and the floor.
The magnitude of the work done by friction can be determined using the formula:
Work done by friction = Force of friction x distance
Since the force of friction is opposing the direction of motion, it is equal to:
Force of friction = mass x acceleration
Now, we can calculate the force of friction:
Force of friction = 3.5 kg x 7.14 m/s^2
Force of friction = 24.99 N
Finally, we can determine the coefficient of kinetic friction. It is defined as the ratio of the force of friction to the normal force (Fn), which is the weight of the book. The normal force is equal to the weight of the book, which is given as 34.3 N:
Coefficient of kinetic friction = Force of friction / Fn
Coefficient of kinetic friction = 24.99 N / 34.3 N
Coefficient of kinetic friction = 0.73
Therefore, the coefficient of kinetic friction between the book and the floor is approximately 0.73.