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Romashka [77]
3 years ago
6

Dwayne ‘The Rock’ Johnson needs to escape from the fourth floor of a burning building (in a movie). He ties a rope around his wa

ist and ties the other end to a 63 kg desk. Then he climbs out of a window and begins to lower himself down. The rope becomes taut when he lets go of the window sill. Then the desk begins to slide on the floor and he starts to accelerate downwards (starting from rest). Dwayne has a mass of 118 kg. The smooth window sill acts as a frictionless pulley. The desk is 10 m from the window. The kinetic friction of the desk sliding on the floor is 370 N. How fast is he travelling when the desk reaches the window? (Hint: Unbeknownst to the audience, The Rock aced Physics 1200 and knows that the tension in the rope is not 370 N).
Physics
1 answer:
ZanzabumX [31]3 years ago
5 0

Answer:

Final Speed of Dwayne 'The Rock' Johnson = 15.812 m/s

Explanation:

Let's start out with finding the force acting downwards because of the mass of 'The Rock':

Dwayne 'The Rock' Johnson: 118kg x 9.81m/s = 1157.58 N

Now the problem also states that the kinetic friction of the desk in this problem is 370 N

Since the pulley is smooth, the weight of Dwayne Johnson being transferred fully, and pulls the desk with a force of 1157.58 N. The frictional force of the desk is resisting this motion by a force of 370 N. Subtracting both forces we get the resultant force on the desk to be: 1157.58 - 370 = 787.58 N

Now lets use F = ma to calculate for the acceleration of the desk:

787.58 = 63 x acceleration

acceleration = 12.501 m/s

Finally, we can use the motion equation:

v^2 - u^2 = 2*a*s

here u = 0 m/s (since initial speed of the desk is 0)

a = 12.501 m/s

and s = 10 m

Solving this we get:

v^2 - 0 = 2 * 12.501 * 10

v = 15.812 m/s

Since the desk and Mr. Dwayne Johnson are connected by a taught rope, they are travelling at the same speed. Thus, Dwayne also travels at            15.812 m/s when the desk reaches the window.

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Answer:

a. wavelength of the sound, \vartheta = 1.315\vartheta_{o}

b. observed frequecy, \lambda = 0.7604\lambda_{o}

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