A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
a. Draw a free body diagram for the textbook.
b. What is the motion in the vertical?
c. What is the net horizontal force acting on the book?
d. What is the net force acting on the book?
e. What is the acceleration of the book?

Respuesta :

Answer:

b) No acceleration in the vertical

c) 35N

d) 35N

e) [tex]8.75\ m/sec^2[/tex]

Explanation:

a) The situation can be shown in the free body diagram shown in the figure below where F is the applied force, Fr is the friction force, W is the weight of the book and N is the normal force exerted vertically up from the desk to the book

b) The vertical movement is restrained by the normal force which opposes to the weight. In absence of any other force, they both are in equilibrium and the net force is zero

c) The net horizontal force acting on the book is the vectorial sum of the applied force and the friction force. Since they lie in the same axis and are opposed to each other:

[tex]Fh_{net}=F-F_r=50 N - 15N=35N[/tex]

d) The net force acting on the book is the vector sum of all forces in all axes. The normal and the weight cancel each other in the y-axis, so our resulting force is the x-axis net force, computed as above:

[tex]F_n=35N[/tex] in the x-axis

e) Following Newton's second law, the acceleration is calculated as

[tex]a=\frac{F_t}{m}=\frac{35N}{4Kg}=8.75m/sec^2[/tex]

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