Can someone help me with part e) ? :) Look at the system in the photo. Object B
ID: 2288046 • Letter: C
Question
Can someone help me with part e) ? :)
Look at the system in the photo. Object B has the mass 10.0 kg.
a) What will the acceleration be if m=4.0 kg and ?k=0.20
b) What is the pulling force in the string in part a) ?
c) what is the acceleration of the system if m=35kg and ?k =0,080?
d) Now the system moves to the left with constant speed when m=28kg. What is the ?k and the pulling force in the force ?
e) Now: m=20 kg and ?s =0,45. An extra mass is loaded on object B untill the system starts to move. How much extra mass is needed? and what will the acceleration be if ?k = 0,37 ?
Explanation / Answer
Solution:
e)
Solution: Let extra mass added = X
Total hanging mass = 10 + X
Weight of the hanging mass = (10+X)g
us= 0.45
Friction force = us mg cos35 = (0.45)(20)g cos35
The force equations: T- us mg cos 35-mg sin35=0 => T = us mg cos35 + mg sin 35 =(0.45)(20)(g)cos35 + (20g)(sin35) =18.4 g newtons
And for the hanging mass: (10+X) g = T
-mg sin35-uk mg cos 35 + T = ma (acceleration is assumed to be in the direction of T, up the ramp)
and -T + (10+X)g =(10+X)a
adding these two equations, - mg sin35-uk mg cos 35 + (10+X)g =20a+(10+8.84)a
=-mg sin35 –ukmg co35 +(10+8.84)g =38.84 a
= -171.8 + 184.65 =38.84a
= 356.5 =38.84a
=>acceleration =a = 356.5/38.84
= 9.18 m/s2
= 9.2 m/s2