A block weighing 10 newtons is resting on a plane inclined 30° to the horizontal. What is the magnitude of the normal force acti ng on the block? Round your answer to the nearest whole number.
       
      
                
     
    
    
    
    1  answer:
            
              
Magnitude of normal force acting on the block is 7 N 
Explanation: 
10N = 1.02kg
Mass of the block = m = 1.02 kg
Angle of incline Θ
=  30°
Normal force acting on the block = N
From the free body diagram,
N = mgCos Θ
N = (1.02)(9.81)Cos(30)
N = 8.66 N
Rounding off to nearest whole number,
N = 7 N
Magnitude of normal force acting on the block = 7 N 
 
                                
             
         	
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