Two 10-kilogram boxes are connected by a massless string thatpasses over a massl
ID: 1675803 • Letter: T
Question
Two 10-kilogram boxes are connected by a massless string thatpasses over a massless frictionless pulley as shown above. Theboxes remain at rest, with the one on the right hanging verticallyand the one on the left 2 meters from the bottom of an inclinedplane that makes an angle of 60 degrees with the horizontal. Thecoefficients of kinetic friction and static friction between theleft-box and the plane are 0.15 and 0.30, respectively. You may useg=10m/s^2, sin 60=0.87 and cos 60=0.50. a. What is the tension T in the string? b.Determine the magnitude of the frictional force acting onthe box on the plane. The string is then cut and left-hand box slides down theinclined plane. c.Determine the amount of mechanical energy that is convertedinto thermal energy during the slide to the bottom. d.Determine the kinetic energy of the left-hand box when itreaches the bottom of the plane.Explanation / Answer
(a) For right hand box T = mg T =10 kg *9.8m/s2 =98 N =98 N (b) For left hand box FN=mg cos =10kg*9.8m/s2*cos60 =49 N Frictional force acting on the boxFs=sFN where coefficient of static friction s=0.30 Fs=0.30*49 =14.7N (d) For left hand box , fromequation of motion v2-u2 =2as v2 -0=2*9.8*2 v2 =39.2m/s Kinetic energy of left hand box=(1/2)mv2 = (1/2)*10*39.2 =196 j = (1/2)*10*39.2 =196 jRelated Questions
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