Calculating Friction Force for Sliding Book

[Fusilli_Jerry89]

I don't know if anyone solves physics problems here, but I havn't done this in over a year and can't concentrate right now so I need help.

A 1.2 kg book is given a shove at 2.0 m/s. It slides across the table coming to a stop ).67 metres later. a) what is the magnitude of the force that stops the book? b) What is the coefficient of kinetic friction between the book and table and c) If released at the same speed, would a heavier book slide the same distance? Explain

for C i think yes because even though there is more force moving, for friction will be acting against it.
for b I got 0.3 by dividing the acceleration(-3.0) by (-9.8), I am unsure how to get a though.

 

[quickslant]

Not 100% sure but i think this is how its done

for part a) first calculate the acceleration the book undergoes
which i get to equal -2.99 m/s^2. The only forces acting on the object is the force of friction whereas gravitational force and normal force cancel out. so Ff = mass * (acceleration) thus (1.2) * (2.99) = 3.59 N

 

[Andrew Mason]

A 1.2 kg book is given a shove at 2.0 m/s. It slides across the table coming to a stop ).67 metres later. a) what is the magnitude of the force that stops the book?

a) You can use energy. What is the kinetic energy of the book after being shoved and traveling at 2.0 m/sec? All of that energy is used doing friction work (ie. KE = friction force x distance).

Or you can simply find the rate of deceleration and multiply by the mass. Your answer of 3.0 m/sec^2 for deceleration is correct.

b) What is the coefficient of kinetic friction between the book and table.
for b I got 0.3 by dividing the acceleration(-3.0) by (-9.8), I am unsure how to get a though

b) You know the magnitude of the friction force from a). How is this force related to the normal force (mg) and the coefficient of kinetic friction \mu_k?

and c) If released at the same speed, would a heavier book slide the same distance? Explain

for C i think yes because even though there is more force moving, for friction will be acting against it.

You have to explain this. You need to show that if the force of friction is proportional to the object's mass, the stopping distance for objects of different mass will be the same. Is the friction force proportional to the object's mass? Why?

AM

 

[edavey8205]

Not 100% sure but i think this is how its done

for part a) first calculate the acceleration the book undergoes
which i get to equal -2.99 m/s^2. The only forces acting on the object is the force of friction whereas gravitational force and normal force cancel out. so Ff = mass * (acceleration) thus (1.2) * (2.99) = 3.59 N

This is right. To calculate the acceleration that he got use v^2 = v0^2 + 2a(x-x0). You know v=2.0 m/s v0 = 0 (initially at rest, the book had to be pushed), a = what you are solving for, and x = 0.67 m x0 = 0. So... a = ?. Then use the equation that he mentioned and you got the force.