Calculate the magnitude of the normal force on a 15.2 kg block in the following circumstances. (Enter your answers in N.) (a) The block is resting on a level surface. ____ N (b) The block is resting on a surface tilted up at a 35.8 angle with respect to the horizontal. N (c) The block is resting on the floor of an elevator that is accelerating upward at 3.53 m/s _ N (d) The block is on a level surface and a force of 155 N is exerted on it at an angle of 35.8 below the horizontal. ___N

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boffeemadrid boffeemadrid · Answer 1 · 2021-01-05T08:33:19+01:00

Answer:

[tex]149.112\ \text{N}[/tex]

[tex]120.94\ \text{N}[/tex]

[tex]302.77\ \text{N}[/tex]

[tex]58.44\ \text{N}[/tex]

Explanation:

m = Mass of block = 15.2 kg

g = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]

[tex]\theta[/tex] = Angle

a = Acceleration

F = Applied force

Normal force is given by

[tex]N=mg\\\Rightarrow N=15.2\times 9.81\\\Rightarrow N=\boldsymbol{149.112\ \mathbf{N}}[/tex]

[tex]\theta=35^{\circ}[/tex]

[tex]N=mg\cos\theta\\\Rightarrow N=15.2\times 9.81\times \cos35.8^{\circ}\\\Rightarrow N=\mathbf{120.94\ N}[/tex]

[tex]a=3.53\ \text{m/s}^2[/tex]

[tex]N=m(g+a)\\\Rightarrow N=15.2(9.81+3.53)\\\Rightarrow N=\boldsymbol{302.77\ \mathbf{N}}[/tex]

[tex]F=155\ \text{N}[/tex]

[tex]N=mg-F\sin35.8^{\circ}\\\Rightarrow N=15.2\times 9.81-155\sin35.8^{\circ}\\\Rightarrow N=\boldsymbol{58.44\ \mathbf{N}}[/tex]