Problem 1:
Constant velocity implies no acceleration. No acceleration implies no net force.
Problem 2:
The woman doubles the applied force, so now there is a net force in the direction of the
applied force. A net force implies acceleration. F = ma.
Problem 3:
The woman exerts no force, so now there is a net force in the opposite direction. A net
force implies acceleration. The acceleration is in the direction opposite to the velocity.
Problem 4:
The block is not accelerating, neither in the vertical nor in the horizontal direction.
(a) The net force in the vertical direction must be zero
(b) The net force in the horizontal direction must be zero.
Let theta be the angle that F makes with the horizontal direction.
Then from (a) W = F sin(theta) + N for the net force in the vertical direction to be zero..
From (b) F cos(theta) = k for the net force in the horizontal direction to be zero.
Problem 5:
The block moves with constant acceleration. Let d denote the distance traveled and a the
acceleration.
d = (1/2)at2, therefore a = 2d/t2 = 3m/(4s2).
F = ma yields the net force.
The net force is F = mg sin (45 deg) – f, where f is the frictional force.
We have f = coefficient * N = mg sin (45 deg) – F.
But N = mg cos(45 deg). Use this in the equation for f and solve for the coefficient.
Problem 6:
Newton’s 3rd law: For every force that an object exerts on a second object, there is a
force equal in magnitude but opposite in direction exerted by the second object on the
first object.