When an object with a mass of 10kg is placed on the horizontal ground, the dynamic friction coefficient between the object and the ground is 0.2 50N, the direction is 37 ° to the horizontal, and the object is pulled upward obliquely for 10m Calculate: the kinetic energy of the object when it moves to 10m position. (g takes 10m / S2; sin 37 ° = 0.6; Cos 37 ° = 0.8) Answer: 260 J To find the process,

The component force F ‖ = fcos37 ° = 0.8 * 50 = 40n in horizontal direction,
The component force of tensile force F along the vertical direction f ⊥ = fsin37 ° = 0.6 * 50 = 30n,
Friction f = u (Mg-F ⊥) = 0.2 * (100-30) = 14N,
From kinetic energy theorem
Ek=W=（F∥-f）S=26*10=260J.

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