Please show all steps of how to get the answer. Thank you A block with mass M 7.

Question

Please show all steps of how to get the answer. Thank you

A block with mass M 7.50 kg is initially moving up the incline and its speed is increasing at a rate of a 4.17 Nm/s2 The applied force Fis horizontal. The coefficients of friction between the block and incline are pus 0.443 and Auk* 0.312. The angle of the incline is 25.0 degrees. .5 kg (a) What is the magnitude of the force F N (t 0.2 N) (b) What is the magnitude of the normal force N between the block and incline N (t 0.2 N) (c) What is the magnitude ofthe force of friction on the block? N (t 0.2 N)

Explanation / Answer

normal force between the incline and box = 7.5g cos25 +F sin25

friction force between the two = ( 7.5g cos25 +F sin25 ) *fric coeff. =  (7.5g cos25 +F sin25 )*0.312

hence : 4.17 * 7.5 = F COS 25 - (7.5g cos25 +F sin25 )*0.312

F = 67.2198 N

PUT THE VALUE OF F IN THE ABOVE EQUATIONS TO GET THE VALUES OF NORMAL AND FRICTION FORCES

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