# A 1.00 × 10º-N crate is being pushed across a level floor at aconstant speed by a force É of 3.00 x 10² N at an angle of 20.0°below the horizontal, as shown in Figure P4.23a. (a) What is thecoefficient of kinetic friction between the crate and the floor?(b) If the 3.00 × 10²N force is instead pulling the block at anangle of 20.0° above the horizontal, as shown in Figure P4.23b,what will be the acceleration of the crate? Assume that the coef-ficient of friction is the same as that found in part (a).20.0°120.0°Figure P4.23

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1 views help_outlineImage TranscriptioncloseA 1.00 × 10º-N crate is being pushed across a level floor at a constant speed by a force É of 3.00 x 10² N at an angle of 20.0° below the horizontal, as shown in Figure P4.23a. (a) What is the coefficient of kinetic friction between the crate and the floor? (b) If the 3.00 × 10²N force is instead pulling the block at an angle of 20.0° above the horizontal, as shown in Figure P4.23b, what will be the acceleration of the crate? Assume that the coef- ficient of friction is the same as that found in part (a). 20.0° 120.0° Figure P4.23 fullscreen
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Step 1 help_outlineImage TranscriptionclosePart (0) F sin (20) F cos(20) mg Since the object is corest moving with constant speed Đ SE=O Efx=0 Fcos (20)–fk = ft= Fcos (20) = (300) cor(20) c 28(-9 N fk=2819 N 281.9 Fojuation O (HE) N = 2819 \$ Mk fullscreen
Step 2 help_outlineImage TranscriptioncloseFor the block, EFfy F sin(20)+mg=N N= 300xsin (20) + 100 = 1l02.6 N From Equation o 2819 = 02557 Mk = (102.6 fullscreen
Step 3 help_outlineImage TranscriptioncloseParet Cbl F sin(20) F Cos(20) bing fk For the block, E fy N+ F sin(20) = mg N= mg - Fsin (20) N= (000 -(300) sin (201 N= 897.39N ew tons. fullscreen

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