a 60.0 n force is applied at an angle of 35 degrees to a 7.00 kg block pressed a

Question

a 60.0 n force is applied at an angle of 35 degrees to a 7.00 kg block pressed against a rough vertical wall and the block slides down the wall at constant velocity. A. What is the coeffiant of kinetic friction between the block and the wall? B. What is the normal force as a function of F? We are looking for the unknown applied force, Using Newton's second law of motion. C. What is the kinetic friction force as a function of F? D. What is the coeffiant of kinetic friction in the vertical direction?

Explanation / Answer

where
? = the angle between the wall and the applied force F.


b) What is the kinetic friction force Fkf as a function of F?

(2) Fkf = Fn * ?k

= F * sin(?) * ?k

where
?k = coefficient of kinetic friction between the block and the wall



c) What is the coefficient of kinetic friction between the block and the wall?

Since the block is sliding at a constant velocity, there is zero net force on the block in the vertical direction. Therefore, the force of gravity on the block Fg is pushing the block down, and it equals Fkf, which is pushing up, and is opposite to it:

(3) Fg = m * g

= 7.00 * 9.81

= 68.7N.

Also,

(4) Kkf = -68.7N

Putting (2) and (4) together:

(5) -68.7 = F * sin(?) * ?k ?

(6) ?k = 68.7 / (F * sin(?) )

= 68.7 / (60.0 * sin(35.0) )

= 68.7 / (60.0 * 0.573 )

= 2.00 Coefficient of friction. That is unusually high, so I will check my work . . .

Well, it seems to check out; so the wall is very rough.

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