A 400 N block is dragged along a horizontal surface by an applied force F→ as shown. The coefficient of kinetic friction is μk=0.4 and the block moves at constant velocity. The magnitude of F→ is:

f=Fcos370μN=4F5μ(Mg−3F5)=4F50.4 Mg−1.2F5=4F50.4 Mg=5.2F54×400=52F5F=200013N=154 N

A 400 N block is dragged along a horizontal surface by an applied force F→ as shown. The coefficient of kinetic friction is μk=0.4 and the block moves at constant velocity. The magnitude of F→ is:

f=Fcos370μN=4F5μ(Mg−3F5)=4F50.4 Mg−1.2F5=4F50.4 Mg=5.2F54×400=52F5F=200013N=154 N

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A 400 N block is dragged along a horizontal surface by an applied force $\vec{F}$ as shown. The coefficient of kinetic friction is $μ_{k} = 0.4$ and the block moves at constant velocity. The magnitude of $\vec{F}$ is:

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100 N

154 N

200 N

290 N

answer is B.

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Detailed Solution

$\begin{array}{l} f = F \cos 37^{0} \\ μ N = \frac{4 F}{5} \\ μ (M g - \frac{3 F}{5}) = \frac{4 F}{5} \\ 0.4 M g - \frac{1.2 F}{5} = \frac{4 F}{5} \\ 0.4 M g = \frac{5.2 F}{5} \\ 4 \times 400 = \frac{52 F}{5} \\ F = \frac{2000}{13} N = 154 N \end{array}$