Answer:
Step one : read the directions
Step two: complete the assignment 
Explanation:
 
        
             
        
        
        
Lets se
And


So

If spring constant is doubled mass must be doubled
 
        
             
        
        
        
B. Aluminum is possibly correct
 
        
             
        
        
        
Answer:
The force is  
Explanation:
From the question we are told that
     The  mass of the block is  
      The  coefficient of  static friction is  
      The coefficient of kinetic friction is  
The  normal force acting on the block is   
       
substituting values 
      
      
Given that the force we are to find is the force required to get the block to start moving then the force acting against this force is the static frictional force which is mathematically evaluated as 
         
substituting values
         
         
Now for this  block to move the force require is  equal to  i.e
 i.e 
        
=>    
         
     
 
        
             
        
        
        
Answer:
a) see attached, a = g sin θ
b)
c)   v = √(2gL (1-cos θ))
Explanation:
 In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
           Wₓ = m a
           W sin θ = m a
           a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
                 θ' = 9/2  θ
              
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
             Em₀ = mg h = mg L (1-cos tea)
Lowest point
           Emf = K = ½ m v²
           Em₀ = Emf
           g L (1-cos θ) = v² / 2
               v = √(2gL (1-cos θ))