A 1,320-N crate is being pushed across a level floor at a constant speed by a fo
ID: 2275411 • Letter: A
Question
A 1,320-N crate is being pushed across a level floor at a constant speed by a force of 290 N at an angle of 20.0 degree below the horizontal, as shown in the figure (a) below. (a) What is the coefficient of kinetic friction between the crate and the floor? ?k = (b) If the 290-N force is instead pulling the block at an angle of 20.0 degree above the horizontal, as shown in the figure (b), what will be the acceleration of the crate? Assume that the coefficient of friction is the same as that found in part (a). m/s2Explanation / Answer
(a) The horizontal component of the 290 N force
= 290 cos 20 = 272.51086 N. This is the force pushing the
crate with uniform velocity and only overcomes friction.
The vertical component of the 290 N force = 290 sin 20 = 99.18584
This adds to the weight of the crate acting vertically
= 99.18584 + 1320 N = 1419.1858 N
Let the coefficient of friction be ? ?? * 1419.1858 N = 272.51086 N
? ? = 272.51086 / 1419.1858 = 0.1920 ? 0.19
(b) In this sitation the vertical component of the applied force
offsets (reduces) the weight of the crate by 99.18584 N .
Hence the frictional force using the above ?
= 0.19 * (1320 - 99.18584) = 231.9547
The excess force after overcoming the friction
= 272.51086 - 231.9547.= 40.5562
This force is now causes the acceleration
= m*a = (1320/9.81)* a = 40.5562 N
? Acceleration of the crate = 40.5562 / (1320/9.81)
= 40.5562 / 134.5565 = 0.3014 ? 0.30 m/sec
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