A force is applied to an initially stationary block of mass 4.00 kg that sits on

Question

A force is applied to an initially stationary block of mass 4.00 kg that sits on a horizontal floor as shown. The 107.0 N force is applied at = 32° angle. The coefficients of friction between the floor and the block are s = 0.506 and k = 0.366. What is the acceleration of the block?

A force is applied to an initially stationary block of mass 4.00 kg that sits on a horizontal floor as shown. The 107.0 N force is applied at e-320 angle. The coefficients of friction between the floor and the block are = 0.506 and k = 0.366, what is the acceleration of the block? m/s2

Explanation / Answer

Normal force , N = m * g + F * cos(theta)

N = 4 * 9.8 + 107 * cos(32 degree)

N = 129.9 N

us = 0.506

uk = 0.366

horizontal componenent of force = 107 * sin(32 degree)
horizontal componenent of force = 56.7 N

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maximum static friction = us * N

maximum static friction = 0.506 * 129.9

maximum static friction = 65.7 N

as the maximum static friction is more than the horizontal component of force

the object will not move

the acceleration of the block is 0 m/s^2

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