Question

The 1.3 kg physics book in figure is connected by a string to a 250 g coffee cup. The book is given a push up the slope and released with a speed of 3.5 m/s . The coefficients of friction are μs =0.50 and μk =0.20.

Part A) How far does the book slide?

An 84.0 kg spacewalking astronaut pushes off a 640 kg satellite, exerting a 95.0 N force for the 0.600 s it takes him to straighten his arms.

Part B) How far apart are the astronaut and the satellite after 1.00 min ?

Answer 1

FBD of cup

ma = mg - T

T = mg - ma

FBD of book

Ma = Mg sin x + u M g cos x

putting value of T

Ma = Mg sin x + mg - ma + u Mg cos x

a ( 1.3 + 0.25) = 1.3* 9.8 sin 20 + 0.25* 9.8 + 0.2* 1.3*9.8* cos 20

a = 5.937 m/s^2

using 3rd equation of motion

v^2 = 2 a d

3.5^2 = 2* 5.937* d

d = 1.03 m

note : sincethe value of angleis not given so i need to take it as 20 degrees, did the same problem 2 days back so i know the aproximation, let me know in comments if it is different than 20deg.

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for astronaut

a = f/m = 95 / 84 = 1.131 m/s^2

velocity in 0.5 s

v1 = at = 1.131 * 0.6 = 0.679 m/s

Distance travelled in next 1 mins

d1 = 0.679* 1.0* 60 = 40.71 m

Distance travelled during push

d2 = 0.5 a ( 0.6)^2

d2 = 0.2034 m

total distance = d1 + d2 = 40.91 m

====

for satellite

a' = 95/ 640 = 0.1484 m/s^2

velocity during push

v = 0.1484* 0.6 = 0.089 m/s

Distance travelled in 1.0 min

d1' = 0.089* 1.0* 60 = 5.3437 m

Distance travelled during acceleration

d2' = 0.5* 0.089* 0.6^2 = 0.016 m

=====

total Seperation

D = d1 + d2 + d1' + d2'

D = 46.27 m

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Comment in case any doubt, will reply for sure.. Goodluck