If 460 kg bear grasping a vertical tree slides down at constant velocity. With acceleration of gravity being 10 m/s^2. What is the fraction force that acts on the bear?
Will I use F=ma, so 460 times 10?
Am I in the right direction
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If 460 kg bear grasping a vertical tree slides down at constant velocity. With acceleration of gravity being 10 m/s^2. What is the fraction force that acts on the bear?
Will I use F=ma, so 460 times 10?
Am I in the right direction
Yes, you are on the right track. If the bear slides at constanjt velocity, then you know that its acceleration is 0. Hence F = 0. This means that the bear's weight is exactly counter-acted by the friction of the bear's claws against the tree. Can you take it from here?
So will the frictional force be 460N
Not quite. The bear's mass is 460 kg, but his weight is a force, given in Newtons. If you know an object's mass, its weight under earth gravity is mg.
Okay I am confused, so exactly what am I doing wrong...
Well... a 460Kg bear has a weight of 460kg * 10m/s^2 = 4600 Newtons.Quote:
Originally Posted by Maria A Gonzale
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