The mass is 3.60kg, T=25N, theta=30 degrees. The surface does have friction, wit
ID: 1546403 • Letter: T
Question
The mass is 3.60kg, T=25N, theta=30 degrees. The surface does have friction, with us=.45 and uk = .39 If x is horizontal and y is vertical. D) what is the friction? G) what is the acceleration of the block? H) what will be the position of the block after 4 seconds if it starts from a position 3m from the left of the origin at a speed of 5 m/s right? I) what will be the velocity of the block at this time? The mass is 3.60kg, T=25N, theta=30 degrees. The surface does have friction, with us=.45 and uk = .39 If x is horizontal and y is vertical. D) what is the friction? G) what is the acceleration of the block? H) what will be the position of the block after 4 seconds if it starts from a position 3m from the left of the origin at a speed of 5 m/s right? I) what will be the velocity of the block at this time? D) what is the friction? G) what is the acceleration of the block? H) what will be the position of the block after 4 seconds if it starts from a position 3m from the left of the origin at a speed of 5 m/s right? I) what will be the velocity of the block at this time?Explanation / Answer
Given that, mass of the block is, m=3.6 kg
tension,T=25 N
theta=30 degrees
static friction is us=0.45
kinematic friction is uk=0.39
D)
The horizontal component of the applied force is.
T*cos(theta)=25*cos 30
= 21.65 N
Normal force, N = m g + T*sin(theta)
= 3.6 kg * 9.8 m/s2 + 25 N * 0.5
= 35.28 + 12.5
= 47.78 N
static friction,
fs = us*N
= 0.45 * 47.78 N
= 21.501 N
Net force = 21.65 N - 21.501 N
= 0.149 N
Kinetic friction fk = uk * N
= 0.39 * 47.78
= 18.634 N
Net force = 21.65 N-18.634 N
= 3.016 N
G)
Fg=mg=3.6 kg*9.8 m/s2
=35.28 N
Fgx=Fgsin(theta)
=35.28 N*sin(30)
=-34.85 N
acceleration of the block is,
a=F/m
=(-34.85 N)/(3.6 kg)
=-9.68 m/s2
H)
v=(h2-h1)/t
h2-h1=v*t
h2=v*t+h1
=5 m/s*4 s+3 m
=23 m
I)
v=distance/time
=23 m/4 s
=5.75 m/s
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