A 70kg object is being pulled up a slope of 30 degrees such that the

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To determine the tension in the rope pulling a 70kg object up a 30-degree slope with a constant velocity, the friction must be considered alongside gravitational forces. The gravitational force acting down the slope is calculated as T = mgsin(30), resulting in 343N without friction. The frictional force, which opposes the motion, is found using the coefficient of friction (0.3) and the normal force. The normal force is calculated as mgcos(30), leading to a frictional force of approximately 0.3 times the normal force. The final tension in the rope must account for both the gravitational pull and the frictional force to maintain constant velocity.
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A 70kg object is being pulled up a slope of 30 degrees such that the rope is parallel to the slope. v is constant. coefficient of friction = 0.3.

what is the tension of the rope?



I know without friction its, something like, T=mgsin(30)=70kg(9.8m/s^2sin(30))= 343N
But i just don't no know to use the friction coefficient? Any help would be much appricated...
 
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