Question

Margaret Mary, Doug , and Kako are at a local amusement park. They see an attraction called the Giant Slide, which is simply a very long and high inclined plane. Visitors at the amusement park climb a long flight of stairs to the top of the 27inclined plane and are given canvas sacks. They sit on the sacks and slide down the plane 70m long. At the time when the three friends walked past the slide, a 135-kg man and a 20-kg boy are each at the top preparing to slide down. "I wonder how much less time it will take the man to slide down than it will take the boy," said Margaret Mary. "I think the boy will take less time," said Doug. "You're both wrong,"said Kako. "They will reach the bottom at the same time."

Answer 1

The question is incomplete. The compete question is :

Margaret Mary, Doug , and Kako are at a local amusement park. They see an attraction called the Giant Slide, which is simply a very long and high inclined plane. Visitors at the amusement park climb a long flight of stairs to the top of the 27inclined plane and are given canvas sacks. They sit on the sacks and slide down the plane 70m long. At the time when the three friends walked past the slide, a 135-kg man and a 20-kg boy are each at the top preparing to slide down. "I wonder how much less time it will take the man to slide down than it will take the boy," said Margaret Mary. "I think the boy will take less time," said Doug. "You're both wrong,"said Kako. "They will reach the bottom at the same time."

If the man and the boy do not take the same amount of time to reach the bottom of the slide, calculate how many seconds of difference there will between the two times.

Solution :

From body sliding down an incline plane, the acceleration is given by,
`$a = g \sin \theta$`

Since the body slides from rest, the initial velocity is, u = 0

Now the equation of motion, we have

v= u + at

or v = 0 +at

or v = at

or
`$t=(v)/(a)$` ............(1)

again we have,
`$v^2 -u^2 =2as $`

or
`$v = √(2s g \sin \theta) $`

Then from equation (1), we have

`$t=(√(2sg \sin \theta))/(g \sin \theta)$`

Thus this means that the time of sliding down an inclined plane does not depend on the mass of the body that slides.

Therefore, Kako is correct.

Since the time taken to slide an inclined plane is same for both the man and the boy, there is no difference between their time of sliding.