A block of mass 1.5 kg is placed on a flat surface, and it is being pulled horiz

Question

A block of mass 1.5 kg is placed on a flat surface, and it is being pulled horizontally by a spring with a spring constant 1.2 times 103 N/m (see Fig. 6.48). The coefficient of static friction between the block and the table is mu1 = 0.60, and the coefficient of sliding friction is muk = 0.40. By what amount must the spring be stretched to start the block moving? What is the acceleration of the block if the stretch of the spring is maintained at a constant value equal to that required to start the motion? By what amount must the spring be stretched to keep the mass moving at constant speed?

Explanation / Answer

Part A: SFx = 0 SFy = 0 N - mg = 0 N = 1.5(9.81) = 14.715 N Ff = cof(s) N = .60*14.715 = 8.829 N 8.829 N / 1200 N/m = .0074m = 7.40mm The spring will have to be pulled to 7.40mm to get the block moving! Part B F= ma 8.829 - .4(14.475)= 1.5 a 2.943 = 1.5a a = 1.962 m/s^2 Part C F - Ff = ma F - .40 (8.829) = 1.5 * 0 F = 3.5316 N 3.5316 N / 1200 N/m = .0029m = 2.90 mm If stretched 2.90mm then the block will have a constant velocity.

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