NEWTON'S LAWS (WITH SURFACE FRICTION)) 5. A mass M = 25 kg is pushed by a constant force, FA = 510 N, acting downward at an angle 0 = 30 degrees from the horizontal as shown. The coefficient of sliding friction between the block and the surface is μ = 0.46. (a) Use Newton's 2nd Law to find the force of friction, FF, acting on the block. (b) FA 0 M Hk=0.56 for the Find the horizontal acceleration of the block. FF=
NEWTON'S LAWS (WITH SURFACE FRICTION)) 5. A mass M = 25 kg is pushed by a constant force, FA = 510 N, acting downward at an angle 0 = 30 degrees from the horizontal as shown. The coefficient of sliding friction between the block and the surface is μ = 0.46. (a) Use Newton's 2nd Law to find the force of friction, FF, acting on the block. (b) FA 0 M Hk=0.56 for the Find the horizontal acceleration of the block. FF=
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Transcribed Image Text:NEWTON'S LAWS (WITH SURFACE FRICTION))
5. A mass M = 25 kg is pushed by a constant force, FA = 510 N, acting downward at
an angle = 30 degrees from the horizontal as shown. The coefficient of sliding friction
between the block and the surface is μ = 0.46.
(a)
Use Newton's 2nd Law to find the force of friction, FF, acting on the block.
(b)
FA
M
Hk=0.56
Perles
ession for the horizontal
Find the horizontal acceleration of the block.
FF=
a=
3, Guad
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