Download Physics – Chp. 6 – Homework p. 136

Physics – Chp. 6 – Homework
p. 136-138 Concept Review # 7, 10, 14, 15, 18
Problems # 1-10
Concept Review
7.
Differences: It takes more force to start an object from rest than to keep an object moving. Thus,
the static coefficient of friction will be higher than the kinetic coefficient of friction.
Similarities: Like all other types of friction, it is a resistive force between two surfaces.
10.
The coefficient of static friction because it takes more force to break the surfaces loose than to
maintain the sliding motion.
14.
a. False
15.
18.
Left
The spring with the spring constant of 10 N/m.
b. False
c. True
Problems
1.
a. F = m*g = (60 kg)(9.8 N/kg) = 588 N
b. F = 39.2 N
c. 690 N
2.
a. m = F/g = (15000 N)/(9.8 N/kg) = 1530.6 kg
b. F = 0.52 kg
3.
This will be the sum of the forces. Force of gravity is always present(mass x 9.8 N/kg), which
will be the force pushing down on the surface, so the force the 75 kg mass “feels” is the equal but
opposite force pushing back up from the surface (Normal force). If the elevator was accelerating
up, that would cause more force to be felt on the 75 kg mass. But in this case, the elevator is
moving with the mass, so the force will be opposite of the weight (force) of the 75 kg mass in the
downward direction.
∑F = F1 + F2 = (75 kg)(9.8 N/kg) – (75 kg)(4.9 m/s2) = 367.5 N
*remember – the units of a Newton are kg*m/s2
4.
*use the coefficient of static friction for rubber on concrete
*find the normal force first, then plug into the friction equation is the coefficient
Fn = m*g = (55 kg)(9.8 N/kg) = 539 N
Ff = µ*Fn = (0.80)(539 N) = 431.2 N
5.
Ff = µ*Fn
200 N = µ*500 N
6.
You push off the wall with 100 N of force, so the wall will push back on you with equal but
opposite force. Make away from the wall the positive direction. The force of friction from your
skates and the ice with oppose (or resist) the force applied from the wall to you (100 N). So,
Fyou – Ff = m*a = 100 N – (0.10)(50 kg*9.8 N/kg) = 50 kg * a
a = 1.02 m/s2
7.
Show arrows pointing in the direction of the forces acting. Artistic talent does not count.
8.
*arbitrarily choose left as positive or right as positive. I made left positive.
∑F = m*a = F1 + F2 = 20 N + (-15 N) = 2.5 kg * a
a = 2 m/s2
9.
*use Hooke’s Law -> F = -k*x
a.
b.
c.
10.
F = -(20 N/m)(0.5 m) = - 10 N
F = -(3 N/m)(0.1 m) = - 0.3 N
F = -(5 x 107)(1 x 10-5 m) = -500 N
a. Should draw four forces: Weight, normal force, force of friction, and force from being struck.
b.
Fn = (0.15 kg)(9.8 N/kg) = 1.47 N
c.
Ff = µ*Fn
0.147 N = µ*(1.47 N)
µ = 0.1