Solution

1

a) Using Newton's Second Law, we get F_net = m a = 12.0 N.
b) If 12.0 N acts upon a mass of 4 kg, the acceleration is 3.0 m/s/s.

2

We can calculate the speed of the ball immediately before striking the floor and immediately upon leaving the floor by using the heights given and the equation v² = 2gh (derived from v_f² - v_o² = 2aDy).

v₁ = - 24.3 m/s
v₂ = + 19.8 m/s.

 

The acceleration provided by the floor is, therefore,

 

Dv/Dt = (19.8 - (-24.3))/.002 = 2.21 x 10⁴m/s²

This gives a force of F_net = mass x acceleration = .5 kg x 2.21 x 10⁴ m/s/s = 1.10 x 10⁴ N in the upward direction.

 

 

3

 

Isolate the two objects and identify the forces acting upon each. Then set the net force on each equal to the mass of the object times its acceleration. For object m₁

T = m₁ a (the weight is balanced by the normal force)

and for object m₂

w₂ - T = m₂ a. (w₂ = m₂g)

Solving, we find:

a = 6.54 m/s/s

and by substituting into either of the two equations

T = 32.7 N.

To see an AVI movie of the problem, click HERE

4

 

 

a) Given v_i = 12.0 m/s and v_f = 6.0 m/s if t = 5 s. Equation #3 gives an acceleration of:
a = - 1.20 m/s/s.
b) If we set the net force equal to the mass times the acceleration we have:
f_k = m a. However f_k = m_kmg in this particular problem. The mass will cancel on both sides of the equation
Solving for m_k we get:
0.122