I\'ve looked up solutions on here to similar problems and tried plugging in my n

Question

I've looked up solutions on here to similar problems and tried plugging in my numbers, but can't get it for some reason still.

Block pulled at an angle A 5.38 kg block located on a horizontal floor is pulled by a cord that exerts a force F-14.9 N at an angle = 19.5° above the horizontal, as shown in the Figure The coefficient of kinetic friction between the block and the floor is 0.100. What is the speed of the block 5.10 s after it starts moving? 13.3 m/s The normal force is NOT just mg in this case! Submit Answer Incorrect. Tries 2/10 Previous Tries

Explanation / Answer

Force acting on the block is given by;

F cos q-m Mg=Ma

14.9 N cos 19.50-0.100* 5.38 kg*9.8 m/s^2=5.38 kg*a

5.38 a=14.04-5.2724

a=1.629 m/s^2

now using d=first equation of motion;

v=ut+(1/2) at2

v=0+(1/2)* 1.629 m/s^2*(5.10 s)2

v=21.185 m/s

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