Chapter 5

Download Document
Showing pages : 1 - 3 of 31
This preview has blurred sections. Sign up to view the full version! View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: CHAPTER 5 Applications of Newtons Laws 1* Various objects lie on the floor of a truck moving along a horizontal road. If the truck accelerates, what force acts on the objects to cause them to accelerate? Force of friction between the objects and the floor of the truck. 2 Any object resting on the floor of a truck will slide if the trucks acceleration is too great. How does the critical acceleration at which a light object slips compare with that at which a much heavier object slips? They are the same. 3 True or false: ( a ) The force of static friction always equals m s F n . ( b ) The force of friction always opposes the motion of an object. ( c ) The force of friction always opposes sliding. ( d ) The force of kinetic friction always equals m k F n . ( a ) False ( b ) True ( c ) True ( d ) True 4 A block of mass m rests on a plane inclined at an angle q with the horizontal. It follows that the coefficient of static friction between the block and the plane is ( a ) m s 1 . ( b ) m s = tan q . ( c ) m s tan q . ( d ) m s tan q . ( d ) 5* A block of mass m is at rest on a plane inclined at angle of 30 o with the horizontal, as in Figure 5-38. Which of the following statements about the force of static friction is true? ( a ) f s > mg ( b ) f s > mg cos 30 o ( c ) f s = mg cos 30 o ( d ) f s = mg sin 30 o ( e ) None of these statements is true. ( d ) f s must equal in magnitude the component of the weight along the plane. 6 A block of mass m slides at constant speed down a plane inclined at an angle q with the horizontal. It follows that ( a ) m k = mg sin q . ( b ) m k = tan q . ( c ) m k = 1 - cos q . ( d ) m k = cos q- sin q . ( a ) Acceleration = 0, therefore f k = mg sin q . With F n = mg cos q , it follows that m k = tan q 7 A block of wood is pulled by a horizontal string across a horizontal surface at constant velocity with a force of 20 N. The coefficient of kinetic friction between the surfaces is 0.3. The force of friction is ( a ) impossible to determine without knowing the mass of the block. ( b ) impossible to determine without knowing the speed of the block. ( c ) 0.3 N. ( d ) 6 N. ( e ) 20 N. ( e ) The net force is zero. 8 A 20-N block rests on a horizontal surface. The coefficients of static and kinetic friction between the surface and the block are m s = 0.8 and m k = 0.6. A horizontal string is attached to the block and a constant tension T is Chapter 5 Applications of Newtons Laws maintained in the string. What is the force of friction acting on the block if ( a ) T = 15 N, or ( b ) T = 20 N. ( a ) If m s mg > 15, then f = f s = 15 N ( b ) T > f s,max ; f = f k = m k mg 0.8 (20 N) = 16 N; f = f s = 15 N f = f k = 0.6 (20 N) = 12 N 9* A block of mass m is pulled at constant velocity across a horizontal surface by a string as in Figure 5-39. The magnitude of the frictional force is ( a ) m k mg . ( b ) T cos q . ( c ) m k ( T- mg ). ( d ) m k T sin q . ( e ) m k ( mg + T sin ...
View Full Document