A 9.0 kg hanging weight is connected by a string over a pulleyto a 5.0 kg block

Question

A 9.0 kg hanging weight is connected by a string over a pulleyto a 5.0 kg block sliding on a flat table. If the coefficientof sliding friction is 0.25, find the tension in the string? A string attached to an airborne kite is maintained at anangle of 45 degrees with the horizontal. If a total of 120 mof string reeled in while bringing the kite back to the ground,what is the horizontal displacement of the kite in the process?(assume the kite string doesn't sage) A hill is 100 m long and makes an angle of 20 degrees with thehorizontgal. As a 50 kg jogger runs up the hill, how muchwork does gravity do on the jogger? Thanks!! A 9.0 kg hanging weight is connected by a string over a pulleyto a 5.0 kg block sliding on a flat table. If the coefficientof sliding friction is 0.25, find the tension in the string? A string attached to an airborne kite is maintained at anangle of 45 degrees with the horizontal. If a total of 120 mof string reeled in while bringing the kite back to the ground,what is the horizontal displacement of the kite in the process?(assume the kite string doesn't sage) A hill is 100 m long and makes an angle of 20 degrees with thehorizontgal. As a 50 kg jogger runs up the hill, how muchwork does gravity do on the jogger? Thanks!!

Explanation / Answer

From the free body diagram we have    For block : 9.0kg    T - mg = ma     T - 0.25*9kg*9.8 = 9a     T - 22.05 = 9a   .......1 For block : 5 kg 5g - T = 5a 5*9.8 - T = 5a 49 - T = 5a    .........2 Adding 1 and 2 we get a = 6.737 m/s2 substituting in 1 T - 22.05 = 9* 6.737 T = 82.68 N substituting in 1 T - 22.05 = 9* 6.737 T = 82.68 N      

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