A force is applied to an initially stationary block of mass 4.60 kg that sits on

Question

A force is applied to an initially stationary block of mass 4.60 kg that sits on a horieontal floor as shown. The 70 3 N force hs appled at 8-32 angle. The coefficients of friction between the floor and the block are p-o.506 and p 0.316. What is the acceleration of the block? MMy Two objects are connected by a light string that passes over a frictionless puilley as shown in the figure below. ms -6.20 kg, m2 -6.20 kg and61 When released from rest, m accelerates downward at 1.262 m/s?. For this to happen, the coefficient of kinetic friction must be and to even begin sliding in the first place the coefficient of static friction must be -2 points Se PSES 6.P062 WI My shows a Ferris wheel that rotates five times each minute. It carries each car around a circle of diameter 18.0 m Top

Explanation / Answer

Considering vertical direction,

N = mg + F cos(theta)

N = (4.60 x 9.8) + (70.3 cos32) = 104.70 N

Static friction Force, Fs = uN = 0.506 x 104.70 = 52.98 N

Ff = 0.316 x 104.70 = 33.09 N

Force along horizontal direction,

Fh = F sin(theta) = 37.25 N

Since, static friction is greater than the applied force,

Hence, block will not move and acceleration will be Zero.

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